I would appreciate it if you could go over my proof.
And $\mathcal{P}(A)$ means power set of A. I thought I should tell you since I may not have the right symbol.
Question:
Suppose F is a family of sets. Prove that there is a unique set A that has the following two properties:
a) $F \subseteq \mathcal{P}(A)$
b) $\forall B (F \subseteq \mathcal{P}(B) \rightarrow A \subseteq B)$
Proof:
Existence
a) Let $A = \cup F$. Let x be arbitrary and assume $x \in F$. Let a be arbitrary and assume $a \in x$ Then, clearly, $a \in \cup F$ by the definition of $\cup F$. In other words, $x \subseteq \cup F$ or $x \in \mathcal{P}(\cup F)$ Therefore, $F \subseteq \mathcal{P}(A)$ when $A = \cup F$
b)Let B be arbitrary and assume $F \subseteq \mathcal{P}(B)$ The assumption can be written as
$F \subseteq \mathcal{P}(B) \leftrightarrow x \in F \rightarrow x \in \mathcal{P}(B) \leftrightarrow x \in F \rightarrow(a\in x \rightarrow a \in B)$
where a and x are arbitrary. Let b be arbitrary and assume $b \in \cup F$. Then, $b \in B$ by the definition of $\cup F$ since there is a set $x_0$ such that if $b \in x_o$, then $b \in B$ as $x \in F \rightarrow(a\in x \rightarrow a \in B)$ Therefore, $\cup F \subseteq B$
Uniqueness
Assume $F \subseteq \mathcal{P}(C)$ and $\forall B(F\subseteq \mathcal{P}(B) \rightarrow C \subseteq B)$, and assume the same for D. D and C are arbitrary.
Since B is arbitrary, the below are true.
$F\subseteq \mathcal{P}(D) \rightarrow C \subseteq D$
$F\subseteq \mathcal{P}(C) \rightarrow D \subseteq C$
Since both $F\subseteq \mathcal{P}(D)$ and $F\subseteq \mathcal{P}(C)$ is true by assumption, $C \subseteq D$ and $D \subseteq C$.
Therefore, $C=D$.
Since $C$ and $D$ were arbitrary, uniqueness is proven.
I think your proof is fine, but could be written more simply.
For example, part b) of the existence proof can be written more succinctly as follows. If $F \subseteq \mathcal{P}(B)$, then every set in the family $F$ is a subset of $B$. Then clearly $\bigcup F$ is also a subset of $B$.