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.
 A: $V$ 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 $a\neq v$ 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 $$d(a,z)<\varepsilon <d(A,V^c)=\inf\{d(x,y):x\in A,~y\in V^c\}$$
if we assume that $z\in V^c$ then $a\in A$ and $z\in V^c$ that is 
$$d(A,V^c)=\inf\{d(x,y):x\in A,~y\in V^c\}<d(a,z)<\varepsilon <d(A,V^c)$$
Contradiction whence, $z\in V$ i.e $B_\varepsilon(a)\subset V$.
A: Suppose it is not true, for every $n$ there exists $x_n\in A$ such that $B(x_n,1/n)$ is not in $V$. Since $A$ is compact. You can extract a convergent sequence $x_{n_k}$ from $(x_n)$, write $x=lim_nx_{n_k}, x_n\in A\subset V$. Since $V$ is open, there exists $c>0$ with $B(x,c)\subset V$. There exists $N$ such that $n_k>N$ implies $x_{n_k}\in B(x,c/4)$. Take $n_k$ such that $n_k>N$ and ${1\over{n_k}}<c/4$. This implies that $B(x_{n_k},{1\over{n_k}})\subset B(x,c)\subset V$. Contradiction.
A: Suppose $X$ is a compact metric space, $A\subseteq X$ is closed, $V\subseteq X$ is open, and $A\subseteq V$. Recall that closed subsets of compact spaces are compact, so $A$ is compact. Further, because $V$ is open and $A\subseteq V$, then for every point $a\in A$, there is some $r_a > 0$ such that $B_{r_a}(a) \subseteq V$. Then $\mathcal{B} = \{ B_{r_a}\}_{r\in A}$ is an open cover of $A$.
Applying the Lebesgue lemma to $\mathcal{B}$ tells us that there is some $\delta > 0 $ such that every subset of $A$ having diameter less that $\delta$ is contained in some member of $\mathcal{B}$. Then let $\epsilon = \delta/4$. Notice that for any $a\in A$, we have diam$(B_{\epsilon}(a)) = 2\epsilon = \delta/2 < \delta$. This implies that $B_{\epsilon}(a)$ must be contained in some set $B_{r_x}(x)\in \mathcal{B}$. But then we know $B_{r_x}(x) \subseteq V$, so it follows that $B_{\epsilon}(a) \subseteq B_{r_x}(x) \subseteq V$.
A: Hint: $A$ and $V^c$ are compact sets and $A \cap  V^{c} = \emptyset$ , so take $\epsilon = d(A, V^c ) > 0. $
A: Let $U=X$ \ $V.$ The cases $A=\phi$ or $U=\phi$ are trivial so suppose $A\ne \phi\ne U.$ 
We can readily show that the function $f(x)=d(x,U)=\inf \{d(x,u):u\in U\}$ is continuous. Since $A$ is compact, its image $f(A)$ is a compact subset of $\Bbb R$ so $f(A)$ has a minimum $M.$ And $M> 0$ because $a\in A\land f(a)=0$ implies $a\in \overline U =U,$ which is absurd. 
Let $\epsilon =M/2.$ 
If $a\in A $ and $u\in B_{\epsilon}(a)\cap  U$ then $0<M\leq d(a,U)\leq d(a,u)<M/2 ,$ which is absurd. So $\forall a\in A\;(B_{\epsilon}(a)\subset V).$  
