Exercise 3.7.1 of "How To Prove It" (Vellemen 2006)

Suppose F is a family of sets. Prove that there is a unique set A that has the following two properties:

(a) $\displaystyle F \subseteq \mathcal{P}(A)$

(b) $\displaystyle \forall B (F \subseteq \mathcal{P}(B) \rightarrow A \subseteq B)$

So what I want to show is:

$\displaystyle \exists !A[F \subseteq \mathcal{P}(A) \land \forall B(F \subseteq \mathcal{P}(B) \rightarrow A \subseteq B)]$

I have the existence part down:

Suppose $\displaystyle A = \cup F$. Then $\displaystyle F \subseteq \mathcal{P}(A)$.

Suppose $\displaystyle y \in F$ and $\displaystyle x \in y$ and $\displaystyle B$ such that $\displaystyle F \subseteq B$.

$\displaystyle A = \cup F$, so clearly $\displaystyle x \in A$. Since $\displaystyle F \subseteq \mathcal{P}(B)$ then it

follows that $\displaystyle x \in B$, therefore $\displaystyle A \subseteq B$. Since $\displaystyle x$, $\displaystyle y$ and

$\displaystyle B$ were arbitrary, $\displaystyle \forall B(F \subseteq \mathcal{P}(B) \rightarrow A \subseteq B)$.

Where I'm having trouble is with the uniqueness proof. I believe that in order to prove uniqueness,

I have to show that

$\displaystyle \forall C \{ [F \subseteq \mathcal{P}(C) \land \forall B(F \subseteq \mathcal{P}(B) \rightarrow C \subseteq B)] \rightarrow C = \cup F \}$

The first half of proving the equality is trivial, that is $\displaystyle F \subseteq \mathcal{P}(C)$ so clearly $\displaystyle \cup F \subseteq C$. It's the

other half that is giving me trouble, that is how do I show that $\displaystyle C \subseteq \cup F$? I can see that it is true but for some reason I can't come up with well reasoned argument as to why.