Let $X$ be a set and let $\mathcal{U}\in \mathcal{P}(\mathcal{P}(X))$.
Since $\mathcal{P}(X)$ is posetal, it suffices to prove the condition $(\star )$. So let $\mathcal{U},\mathcal{V}\in \mathcal{P}(\mathcal{P}(X))$ with $\mathcal{U}\subset \mathcal{V}$. We claim that
\[ \bigcup _{U\in \mathcal{U}}U\subset \bigcup _{V\in \mathcal{V}}V. \]
Indeed, given $x\in \bigcup _{U\in \mathcal{U}}U$, there exists some $U\in \mathcal{U}$ such that $x\in U$, but since $\mathcal{U}\subset \mathcal{V}$, we have $U\in \mathcal{V}$ as well, and thus $x\in \bigcup _{V\in \mathcal{V}}V$, which gives our desired inclusion.
We have
\begin{align*} \bigcup _{U\in {\scriptsize \displaystyle \bigcup _{A\in \mathcal{A}}A}}U & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some $U\in \displaystyle \bigcup _{A\in \mathcal{A}}A$}\\ & \text{such that we have $x\in U$} \end{aligned} \right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some $A\in \mathcal{A}$}\\ & \text{and some $U\in A$ such that}\\ & \text{we have $x\in U$} \end{aligned} \right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some $A\in \mathcal{A}$}\\ & \text{such that we have $x\in \bigcup _{U\in A}U$} \end{aligned} \right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\bigcup _{A\in \mathcal{A}}\left(\bigcup _{U\in A}U\right). \end{align*}
This finishes the proof.
We have
\begin{align*} \bigcup _{V\in \left\{ U\right\} }V & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some $V\in \left\{ U\right\} $}\\ & \text{such that we have $x\in U$} \end{aligned} \right\} \\ & = \left\{ x\in X\ \middle |\ x\in U \right\} \\ & = U.\end{align*}
This finishes the proof.
We have
\begin{align*} \bigcup _{\left\{ u\right\} \in \chi _{X}(U)}\left\{ u\right\} & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some $\left\{ u\right\} \in \chi _{X}(U)$}\\ & \text{such that we have $x\in \left\{ u\right\} $} \end{aligned} \right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some $\left\{ u\right\} \in \chi _{X}(U)$}\\ & \text{such that we have $x=u$} \end{aligned} \right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some $u\in U$}\\ & \text{such that we have $x=u$} \end{aligned} \right\} \\ & = \left\{ x\in X\ \middle |\ x\in U\right\} \\ & = U.\end{align*}
This finishes the proof.
Item 5: Interaction With Unions I
We have
\begin{align*} \bigcup _{W\in \mathcal{U}\cup \mathcal{V}}W & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some $W\in \mathcal{U}\cup \mathcal{V}$}\\ & \text{such that we have $x\in W$} \end{aligned} \right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some $W\in \mathcal{U}$ or some}\\ & \text{$W\in \mathcal{V}$ such that we have $x\in W$} \end{aligned} \right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some $W\in \mathcal{U}$}\\ & \text{such that we have $x\in W$} \end{aligned} \right\} \\ & \phantom{={}}\mkern 4mu\cup \left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some $W\in \mathcal{V}$}\\ & \text{such that we have $x\in W$} \end{aligned} \right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left(\bigcup _{W\in \mathcal{U}}W\right)\cup \left(\bigcup _{W\in \mathcal{V}}W\right)\\ & = \left(\bigcup _{U\in \mathcal{U}}U\right)\cup \left(\bigcup _{V\in \mathcal{V}}V\right). \end{align*}
This finishes the proof.
Item 6: Interaction With Unions II
Assume $\mathcal V$ is nonempty. We have
\begin{align*} U \cup \bigcup _{V\in \mathcal{V}}V & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ x\in X\ \middle |\ \begin{aligned} & \text{$x \in U$ or $x \in \bigcup _{V\in \mathcal{V}}V$} \end{aligned} \right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{$x \in U$ or there exists some}\\ & \text{$V\in \mathcal{V}$ such that $x\in V$} \end{aligned} \right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some $V\in \mathcal{V}$}\\ & \text{such that $x \in U$ or $x\in V$} \end{aligned} \right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some $V\in \mathcal{V}$}\\ & \text{such that $x \in U \cup V$} \end{aligned} \right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\bigcup _{V\in \mathcal{V}} U \cup V. \end{align*}
This concludes the proof of the first statement. For the second statement, use Item 4 of Proposition 4.3.8.1.2 to rewrite
\begin{align*} \left(\bigcup _{U\in \mathcal{U}}U\right)\cup V & = V \cup \left(\bigcup _{U\in \mathcal{U}}U\right),\\ \bigcup _{U\in \mathcal{U}}(U\cup V) & = \bigcup _{U\in \mathcal{U}}(V\cup U). \end{align*}
But these two sets are equal by the first statement.
Item 7: Interaction With Intersections I
We have
\begin{align*} \bigcup _{W\in \mathcal{U}\cap \mathcal{V}}W & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some $W\in \mathcal{U}\cap \mathcal{V}$}\\ & \text{such that we have $x\in W$} \end{aligned} \right\} \\ & \subset \left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some $U\in \mathcal{U}$ and some $V\in \mathcal{V}$}\\ & \text{such that we have $x\in U$ and $x\in V$} \end{aligned} \right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some $U\in \mathcal{U}$}\\ & \text{such that we have $x\in U$} \end{aligned} \right\} \\ & \phantom{={}}\mkern 4mu\cup \left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some $V\in \mathcal{V}$}\\ & \text{such that we have $x\in V$} \end{aligned} \right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left(\bigcup _{U\in \mathcal{U}}U\right)\cap \left(\bigcup _{V\in \mathcal{V}}V\right).\end{align*}
This finishes the proof.
Item 8: Interaction With Intersections II
We have
\begin{align*} U \cap \bigcup _{V\in \mathcal{V}}V & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ x\in X\ \middle |\ \begin{aligned} & \text{$x \in U$ and $x \in \bigcup _{V\in \mathcal{V}}V$} \end{aligned} \right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{$x \in U$ and there exists some}\\ & \text{$V\in \mathcal{V}$ such that $x\in V$} \end{aligned} \right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some $V\in \mathcal{V}$}\\ & \text{such that $x \in U$ and $x\in V$} \end{aligned} \right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some $V\in \mathcal{V}$}\\ & \text{such that $x \in U \cap V$} \end{aligned} \right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\bigcup _{V\in \mathcal{V}} U \cap V. \end{align*}
This concludes the proof of the first statement. For the second statement, use Item 5 of Proposition 4.3.9.1.2 to rewrite
\begin{align*} \left(\bigcup _{U\in \mathcal{U}}U\right)\cap V & = V \cap \left(\bigcup _{U\in \mathcal{U}}U\right),\\ \bigcup _{U\in \mathcal{U}}(U\cap V) & = \bigcup _{U\in \mathcal{U}}(V\cap U). \end{align*}
But these two sets are equal by the first statement.
Item 9: Interaction With Differences
Let $X=\left\{ 0,1\right\} $, let $\mathcal{U}=\left\{ \left\{ 0,1\right\} \right\} $, and let $\mathcal{V}=\left\{ \left\{ 0\right\} \right\} $. We have
\begin{align*} \bigcup _{W\in \mathcal{U}\setminus \mathcal{V}}U & = \bigcup _{W\in \left\{ \left\{ 0,1\right\} \right\} }W\\ & = \left\{ 0,1\right\} , \end{align*}
whereas
\begin{align*} \left(\bigcup _{U\in \mathcal{U}}U\right)\setminus \left(\bigcup _{V\in \mathcal{V}}V\right) & = \left\{ 0,1\right\} \setminus \left\{ 0\right\} \\ & = \left\{ 1\right\} . \end{align*}
Thus we have
\[ \bigcup _{W\in \mathcal{U}\setminus \mathcal{V}}W=\left\{ 0,1\right\} \neq \left\{ 1\right\} =\left(\bigcup _{U\in \mathcal{U}}U\right)\setminus \left(\bigcup _{V\in \mathcal{V}}V\right). \]
This finishes the proof.
Item 10: Interaction With Complements I
Let $X=\left\{ 0,1\right\} $ and let $\mathcal{U}=\left\{ 0\right\} $. We have
\begin{align*} \bigcup _{U\in \mathcal{U}^{\textsf{c}}}U & = \bigcup _{U\in \left\{ \text{Ø},\left\{ 1\right\} ,\left\{ 0,1\right\} \right\} }U\\ & = \left\{ 0,1\right\} , \end{align*}
whereas
\begin{align*} \bigcup _{U\in \mathcal{U}}U^{\textsf{c}} & = \left\{ 0\right\} ^{\textsf{c}}\\ & = \left\{ 1\right\} . \end{align*}
Thus we have
\[ \bigcup _{U\in \mathcal{U}^{\textsf{c}}}U=\left\{ 0,1\right\} \neq \left\{ 1\right\} =\bigcup _{U\in \mathcal{U}}U^{\textsf{c}}. \]
This finishes the proof.
Item 11: Interaction With Complements II
We have
\begin{align*} \left( \bigcup _{U \in \mathcal U} U \right)^{\textsf{c}} & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists no $U\in \mathcal{U}$}\\ & \text{such that we have $x\in U$}\end{aligned} \right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{for all $U\in \mathcal{U}$}\\ & \text{we have $x\not\in U$}\end{aligned} \right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ x\in X\ \middle |\ \begin{aligned} & \text{for all $U\in \mathcal{U}$}\\ & \text{we have $x\in U^{\textsf{c}}$}\end{aligned} \right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\bigcap _{U \in \mathcal U} U^{\textsf{c}}. \end{align*}
Item 12: Interaction With Complements III
By Item 11 Item 3 of Proposition 4.3.11.1.2, we have
\begin{align*} \left( \bigcap _{U \in \mathcal U} U \right)^{\textsf{c}} & = \left( \bigcap _{U \in \mathcal U} (U^{\textsf{c}})^{\textsf{c}} \right)^{\textsf{c}}\\ & = \left(\left( \bigcup _{U \in \mathcal U} U^{\textsf{c}} \right)^{\textsf{c}}\right)^{\textsf{c}}\\ & = \bigcup _{U \in \mathcal U} U^{\textsf{c}}. \end{align*}
Item 13: Interaction With Symmetric Differences
Let $X=\left\{ 0,1\right\} $, let $\mathcal{U}=\left\{ \left\{ 0,1\right\} \right\} $, and let $\mathcal{V}=\left\{ \left\{ 0\right\} ,\left\{ 0,1\right\} \right\} $. We have
\begin{align*} \bigcup _{W\in \mathcal{U}\mathbin {\triangle }\mathcal{V}}W & = \bigcup _{W\in \left\{ \left\{ 0\right\} \right\} }W\\ & = \left\{ 0\right\} , \end{align*}
whereas
\begin{align*} \left(\bigcup _{U\in \mathcal{U}}U\right)\mathbin {\triangle }\left(\bigcup _{V\in \mathcal{V}}V\right) & = \left\{ 0,1\right\} \mathbin {\triangle }\left\{ 0,1\right\} \\ & = \text{Ø}, \end{align*}
Thus we have
\[ \bigcup _{W\in \mathcal{U}\mathbin {\triangle }\mathcal{V}}W=\left\{ 0\right\} \neq \text{Ø}=\left(\bigcup _{U\in \mathcal{U}}U\right)\mathbin {\triangle }\left(\bigcup _{V\in \mathcal{V}}V\right). \]
This finishes the proof.
Item 14: Interaction With Internal Homs I
This is a repetition of Item 7 of Proposition 4.4.7.1.3 and is proved there.
Item 15: Interaction With Internal Homs II
This is a repetition of Item 8 of Proposition 4.4.7.1.3 and is proved there.
Item 16: Interaction With Internal Homs III
This is a repetition of Item 9 of Proposition 4.4.7.1.3 and is proved there.
Item 17: Interaction With Direct Images
This is a repetition of Item 3 of Proposition 4.6.1.1.5 and is proved there.
Item 18: Interaction With Inverse Images
This is a repetition of Item 3 of Proposition 4.6.2.1.3 and is proved there.
Item 19: Interaction With Codirect Images
This is a repetition of Item 3 of Proposition 4.6.3.1.7 and is proved there.
Item 20: Interaction With Intersections of Families I
We have
\begin{align*} \bigcap _{U\in {\scriptsize \displaystyle \bigcup _{A\in \mathcal{A}}A}}U & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ x\in X\ \middle |\ \begin{aligned} & \text{for each $U\in \bigcup _{A\in \mathcal{A}}A$,}\\ & \text{we have $x\in U$} \end{aligned} \right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{for each $A\in \mathcal{A}$ and each}\\ & \text{$U\in A$, we have $x\in U$} \end{aligned} \right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\bigcap _{A\in \mathcal{A}}\left(\bigcap _{U\in A}U\right).\end{align*}
This finishes the proof.
Item 21: Interaction With Intersections of Families II
Omitted.