Closed subset of compact metric space contained in an open set I am stuck on the following problem: Suppose $X$ is a compact metric space, $A\subseteq X$ is closed, $V\subseteq X$ is open, and $A\subseteq V$. Show that there exists some $\epsilon > 0$ such that for all $a\in A$, $B_{\epsilon}(a)\subseteq V$. We are given the hint "Lebesgue lemma", but I'm not sure if this means to apply Lebesgue's Lemma, or to use a proof strategy similar to that of the Lebesgue Lemma. My thinking was that if we could determine the shortest distance between any point in $A$ and the boundary of $V$, then we could make that our $\epsilon$. But I don't know how I would make that precise. Also I am not sure whether there actually is a minimum distance. Edit: Ok I believe I have come up with a solution based on Robert Thingum's hint. I will post it below. Let me know if you spot any mistake.

$V$ open implies that $V^c$ is closed , hence $A$ and $V^c$ are compact set such that $ A \cap V^c =\emptyset $ since,

$$A\subset V\implies A \cap V^c =\emptyset$$

therefore their exist $$a\in A ~~~and ~~~v\in V^c~~~\text{such that, }~~d(a,v )=d(A,V^c) =\inf\\{d(x,y):x\in A,~y\in V^c\\}$$

$a\in A ~~~and ~~~v\in V^c$ with $A \cap V^c =\emptyset$ means that $a\
eq v$ thus $$d(A,V^c)=d(a,v )>0 $$

> **Claim** For any $0<\varepsilon <(A,V^c)$ and any $a\in A$ wehave $$B_\varepsilon(a)\subset V$$

**Proof** let $z\in B_\varepsilon(a) $ i.e $$d(a,z)<\varepsilon
if we assume that $z\in V^c$ then $a\in A$ and $z\in V^c$ that is

$$d(A,V^c)=\inf\\{d(x,y):x\in A,~y\in V^c\\}