If $C \subset U$ is compact, $U$ open with compact closure in a metric space $M$, then there exist open $V$ st $C \subset V \subset \bar V \subset U$

Question

Show that if $C \subset U$ is compact, $U$ open with compact closure in a metric space $M$, then there exists open set $V$ such that $C \subset V \subset \overline V \subset U$

Thoughts:

Since $C$ is compact in a metric space, therefore it is closed and bounded. Pick some balls $\mathcal{B}_\epsilon(x)$ for all $x \in \partial C$ such that $\epsilon$ is small enough so $\bigcup_{x \in \partial C} \mathcal{B}_\epsilon(x) \cup \text{int}(C) \subset U$, then construct a $V$ in between these two sets....not sure how to continue.

Answer 1

Let the metric be $d$. Consider $d(C,\partial U)$. Since $C$ is compact and $\partial U$ is closed, $d(C,\partial U)=d(x,y)$ for some $x\in C,\ y\in\partial U$. And since $C\subset U$ with $U$ open, we have $$d_0=d(C,\partial U)=d(x,y)>0$$ Let $V=\{x\in M:d(x,C)<d_0/2\}$.