Is this proof correct for : Does $F(A)\cap F(B)\subseteq F(A\cap B) $ for all functions $F$?

Your gut is right. Something is indeed very wrong: the statement that you’re trying to prove is false. Here’s a familiar counterexample: let $F:\Bbb R\to\Bbb R:x\mapsto x^2$, let $A$ be the set of negative real numbers, and let $B$ be the set of positive real numbers. Then $F[A]=F[B]=B$, so $F[A]\cap F[B]=B$, but $A\cap B=\varnothing$, so $F[A\cap B]=\varnothing$. Clearly $B\nsubseteq\varnothing$!

The problem with your reasoning, as you can see from the counterexample, is that although $x\in F[A]$ guarantees that $x=F(a)$ for some $a\in A$, and $x\in F[B]$ guarantees that $x=F(b)$ for some $b\in B$, you have no guarantee that $a=b$. There may be nothing at all in $A\cap B$.

The third line is mistaken. You only know that there exists an $x$ in $A$ such that $F(x)=y$, and you know there is a $z\in B$ such that $F(z)=y$.

It is extremely easy to find a counterexample: just draw two sets $A$, $B$ that are disjoint, and map an $a\in A$ and a $b\in B$ to a single point. Then you have that $y\in F(A)\cap F(B)$, but $F(A\cap B)=\emptyset$.