Show that there exists $\epsilon >0$ such that $\bigcup_{x\in A}B(x;\epsilon)\subset V.$ Let $X$ be a compact metric space, $A$ a closed subset of $X$ and $V$ an open subset of $X$. Suppose $A\subset V$. Show that there exists $\epsilon >0$ such that $$\bigcup_{x\in A}B(x;\epsilon)\subset V.$$
My approach is as follows. Since $A$ is closed in a compact space $X$, $A$ is compact. But then I am lost, how to continue? Any comment or recommendation is welcome.
 A: $A\times (X\setminus V)$ is compact, hence $(x,y)\mapsto d(x,y)$ assumes its minimum, which must be positive
A: Note that the statement is equivalent to saying that the distance of 
Consider the function $f: x\to dist(x,V^c):=\inf_{y\in V^c}{d(x,y)}$. 
We can prove that this function is continuous fairly easily, and that $f(x)>0 \quad \forall x\in A$.
Now we have a continuous function on a compact set, hence it must assume its minimum, which must, since $f$ is positive, be greater than an $\varepsilon>0$. Now try to work with that, can you show that for this $\varepsilon$, your statement holds?
A: For each $x \in A$, pick $r_x >0$ such that $B(x, 2r_x) \subseteq V$. As $A$ is compact, there are finitely many of the balls $B(x, r_x)$ that cover $A$. Let $\epsilon$ be the minimum of all the finitely many $r_x$.
Now, if $x \in A$, there is some $y$ and $r_y$ such that $x \in B(y,r_y)$ and this ball is one of the finitely many from the subcover above. If $z \in B(x,\epsilon)$, then $$d(z,y) \le d(z, x) + d(y,x) < \epsilon + r_y \le 2r_y$$
This implies that $z \in B(y, 2r_y) \subseteq V$. As $z \in B(x, \epsilon)$ was arbitrary, $\epsilon$ is as required.  
