Let $E,F$ be topological vector spaces, where the topology on $E$ is induced by a metric $d$. Let $\phi:E\to F$ be a continuous, open, surjective map. For any two closed subsets $A,B\subseteq F$ define $$D(A,B):=\inf_{a\in E: \phi(a)=A} d(a,\phi^{-1}(B)).$$ If $D(A,B)=0$, then there must be some $a\in \phi^{-1}(A)\cap \phi^{-1}(B)$. This would come down to finding $a\in E$ such that $\phi(a)\in A$ and $\exists b_n\in E, \phi(b_n)\in B$ with $b_n\to a$ (since $\phi$ is continuous and $B$ is closed). Of course, we could switch the roles of $A$ and $B$.
If $D(A,B)=0$, then $\forall n\exists a_n\in E, \phi(a_n)\in A: d(a_n,\phi^{-1}(B))<1/n$. The candidate $a\in E$ would have to be the limit of $(a_n)_n$, but are we even sure this limit exists?
How can I prove the existence of such an $a$?
This is not true (if I understood your claim correctly). Take $E=F=\mathbb R^2$ and $\phi=id$, and define $$ A=\{(x,y): \ xy=1\}, B=\{(x,y): \ xy=0\}. $$ Then $D(A,B)=0$ but $A\cap B=\emptyset$.
The problem in your attempt is: $d(a_n,B)<1/n$ implies existence of $b_n\in B$ such that $d(a_n,b_n)<1/n$. But neither of the sequences $(a_n)$ or $(b_n)$ needs to be convergent. Here, compactness of one of the sets is needed.