This is exercise $3.7.1$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:
Suppose $\mathcal F$ is a family of sets. Prove that there is a unique set $A$ that has the following two properties:
$(a)$ $\mathcal F\subseteq \mathscr P(A)$.
$(b)$ $\forall B\Bigr(\mathcal F\subseteq \mathscr P(B)\rightarrow A\subseteq B\Bigr)$.
Here is my proof:
Existence: Let $\mathcal F$ be an arbitrary family of sets and suppose $A=\bigcup\mathcal F$.
Let $C$ be an arbitrary element of $\mathcal F$. Let $x$ be an arbitrary element of $C$. From $C\in\mathcal F$ and $x\in C$, $x\in\bigcup\mathcal F$ and since $A=\bigcup\mathcal F$, $x\in A$. Thus if $x\in C$ then $x\in A$. Since $x$ is arbitrary, $\forall x(x\in C\rightarrow x\in A)$ and so $C\subseteq A$. Ergo $C\in\mathscr P(A)$. Therefore if $C\in\mathcal F$ then $C\in\mathscr P(A)$. Since $C$ is arbitrary, $\forall C\Bigr(C\in\mathcal F\rightarrow C\in\mathscr P(A)\Bigr)$ and so $\mathcal F\subseteq\mathscr P(A)$.
Let $B$ be an arbitrary set such that $\mathcal F\subseteq \mathscr P(B)$. Let $x$ be an arbitrary element of $A$. From $\mathcal F\subseteq \mathscr P(B)$ by definition we have $\bigcup\mathcal F\subseteq B$. Since $A=\bigcup \mathcal F$, from $\bigcup\mathcal F\subseteq B$ and $x\in A$ we obtain $x\in B$. Thus if $x\in A$ then $x\in B$. Since $x$ is arbitrary, $\forall x(x\in A\rightarrow x\in B)$ and so $A\subseteq B$. Therefore if $\mathcal F\subseteq \mathscr P(B)$ then $A\subseteq B$. Since $B$ is arbitrary, $\forall B\Bigr(\mathcal F\subseteq\mathscr P(B)\rightarrow A\subseteq B\Bigr)$.
Uniqueness: Let $C$ and $D$ be arbitrary sets such that $\mathcal F\subseteq\mathscr P(C)$,
$\mathcal F\subseteq\mathscr P(D)$, $\forall B\Bigr(\mathcal F\subseteq\mathscr P(B)\rightarrow C\subseteq B\Bigr)$, and $\forall B\Bigr(\mathcal F\subseteq\mathscr P(B)\rightarrow D\subseteq B\Bigr)$. From $\mathcal F\subseteq\mathscr P(C)$ and $\forall B\Bigr(\mathcal F\subseteq\mathscr P(B)\rightarrow D\subseteq B\Bigr)$, $D\subseteq C$. From $\mathcal F\subseteq\mathscr P(D)$ and $\forall B\Bigr(\mathcal F\subseteq\mathscr P(B)\rightarrow C\subseteq B\Bigr)$, $C\subseteq D$. Ergo $C=D$.
$Q.E.D.$
Is my proof valid$?$
Thanks for your attention.
Best Answer
(Your proof looks fine to me. This is strictly speaking not an answer to your question, but perhaps this alternative proof helps provide some insight.)$% \require{begingroup} \begingroup \newcommand{\calc}{\begin{align} \quad &} \newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}} \newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} } \newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\Ref}[1]{\text{(#1)}} \newcommand{\then}{\rightarrow} \newcommand{\when}{\leftarrow} \newcommand{\fa}[2]{\forall #1 \left( #2 \right) } \newcommand{\ex}[2]{\exists #1 \left( #2 \right) } \newcommand{\exun}[2]{\exists ! #1 \left( #2 \right) } \newcommand{\F}{\mathcal F} \newcommand{\equiv}{\leftrightarrow} %$
It seems the key insight here is what $\;\F \subseteq \mathscr P(X)\;$ means, and how we can 'pull out' $\;X\;$ from this expression. The simplest way is to just start calculating, by first expanding the definitions:
$$\calc \F \subseteq \mathscr P(X) \op\equiv\hint{expand definition of subset} \fa D {D \in \F \then D \in \mathscr P(X)} \op\equiv\hint{expand definition of powerset} \fa D {D \in \F \then D \subseteq X} \op\equiv\hint{expand definition of subset} \fa D {D \in \F \then \fa z {z \in D \then z \in X}} \op\equiv\hint{logic: merge quantifications} \fa {D,z} {D \in \F \land z \in D \;\then\; z \in X} \tag{*} \op\equiv\hints{logic -- to simplify in a different way,}\hints{by bringing quantification over $\;D\;$ closer to where it is used,}\hint{keeping $\;X\;$ separate} \fa z {\ex D {D \in \F \land z \in D} \;\then\; z \in X} \op\equiv\hint{definition of union -- to simplify} \fa z {z \in \bigcup \F \;\then\; z \in X} \op\equiv\hint{definition of subset -- to simplify} \bigcup \F \subseteq X \endcalc$$
(Note how $\Ref{*}$ is like a central hinge, connecting the top part which looks at the sets, and the bottom part which looks at the elements in those sets.)
So we discovered that $\;\F \subseteq \mathscr P(X) \;\equiv\; \bigcup \F \subseteq X\;$, and that now makes it simple to prove the statement:
$$\calc \exun A {\F \subseteq \mathscr P(A) \;\land\; \fa B {\F \subseteq \mathscr P(B) \then A \subseteq B}} \op\equiv\hint{using the above property, twice} \exun A {\bigcup \F \subseteq A \;\land\; \fa B {\bigcup \F \subseteq B \then A \subseteq B}} \op\equiv\hint{set theory: basic property of $\;\subseteq\;$, i.e., transitivity} \exun A {\bigcup \F \subseteq A \;\land\; A \subseteq \bigcup \F} \op\equiv\hint{set theory: simplify} \exun A {A = \bigcup \F} \endcalc$$
which is trivially true, and therefore the $\;A\;$ that we were looking for turns out to be $\;\bigcup \F\;$.
$% \endgroup %$