Suppose $x\in \cup F$, then $\forall A \in F (x\in A)$. Since $\forall A \in F (A\subseteq B)$ , $x\in B$.
Is this correct? Its obvious that for any element $x \in \cup F$ , x will be in some $A\in F$, which in turn, $x\in B$ since any $A \subseteq B$.
Contradicting what has been said in the comments, your answer is not (totally) correct.
Just like the OP mentioned in the comments, he defines $\cup F = \cup_{A \in F} A.$ Therefore, when you suppose $x \in \cup F$, one can't guaratee that $x \in A,$ for every set $A \in F$ (this is something the OP stated to be true). What we can do is the following:
Since $x \in \cup F,$ by definition of union of sets we can guarantee that there exists at least one set $A \in F$ such that $x \in A$ (note that this is not necessarily true for every set $A \in F$).
The rest of the resolution is correct and remains the same after this correction.