Let $(\Omega,d)$ be a metric space, $\tau:\Omega\to\Omega$, $A\subseteq\Omega$ be compact and $$B:=\{x\in\Omega:(\tau^n(x))_{n\in\mathbb N}\text{ is eventually in every neighborhood of }A\}.$$
How can we show that $B=\tilde B:=\{x\in\Omega:d(\tau^n(x),A)\xrightarrow{n\to\infty}0\}$?
In general, I know that $d(x,C)=0$ iff $x\in\overline C$. Now the crucial point must be the compactness of $A$; but how can we utilize it?
If $x\in B$ and $\varepsilon>0$, then $N:=\{x\in\Omega:d(x,A)<\varepsilon\}$ is clearly a neighborhood of $N$. So, it holds $B\subseteq\tilde B$; even when $A$ is not compact.
But how can we show the other inclusion? My guess is that we need the compactness of $A$ to ensure (can we?) that if $N$ is a neighborhood of $A$, there is a $\varepsilon>0$ with $A\subseteq O:=\{x\in\Omega:d(x,A)<\varepsilon\}\subseteq N$. This would be impossible if $A$ is not bounded (but I'm unsure why it would fail if it is not closed as well).
Lemma
If $A$ is compact and $A \subseteq O$ where $O$ is open, then there is some $r>0$ such that $\{y \in X\mid \exists a \in A : d(a,y) < r\} \subseteq O$.
This is quite straightforward from the open cover definition of compactness: for each $a \in A$, there is some $r(a)>0$ such that $B(a,2r(a)) \subseteq O$. Then $\{ B(a, r(a))\mid a \in A\}$ is an open cover of $A$ so finitely many, say $B(a_1, r(a_1)), \ldots, B(a_n, r(a_n))$ also cover $A$. If now $a \in A$ and $d(y,a) < r= \min_{i=1}^n r(a_i)$, then $a \in B(a_j, r(a_j))$ for some $1 \le j \le n$ and then $d(y,a(j)) \le d(y, a) + d(a(j),a) < r(a(j)) + r \le 2r(a(j))$ so that $y \in O$ by how the balls were chosen.
I think this little lemma will allow you to show the required equivalence.