Show that there exist a nonempty open set $V$ such that $\overline V\subset U,\overline V\cap A=\emptyset$ in a compact metric space

Let $Y$ be a compact metric space and $A$ be closed in $Y$ and $A^\circ=\emptyset$. Show that if $U$ is a nonempty open set in $Y$, there exist a nonempty open set $V$ such that $\overline V\subset U,\overline V\cap A=\emptyset$.

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HINT: Let $W=V\setminus A$. Show that $W$ is a non-empty open set in $Y$. Let $p\in W$ be arbitrary. Use the fact that a metric space is regular. (Note that $Y$ need not be compact.)
As $A$ has no inner points, but $U$ has, there is a $x \in U \setminus A$. The map $f\colon A \to [0,\infty)$, $a \mapsto d(a,x)$ is continuous, hence attains its minimum $\epsilon > 0$ on $A$. Now let $\delta < \frac \epsilon 2$ be such that the closed ball $\bar B_{\delta}(x)$ is contained in $U$. Set $V := B_\delta(x)$ (the open ball). Then $V$ is non-empty (it contains $x$), open, its closure is contained in $\bar B_\delta(x)$, hence in $U$ and doesn't meet $A$ as all its points have distance at least $\frac \epsilon 2$ from $A$.