Compactness used to get a covering by special smaller balls Suppose $(X,d)$ is a compact metric space. Suppose we have a set $A \subseteq X$ such that the set of open $\epsilon$-balls around the points of $A$ cover $X$. I've read that "By compactness, there exists $\epsilon_{1} < \epsilon$ such that the $\epsilon_{1}$-balls centered at the points of $A$ also cover $X$." Do any of you have hints for how to think about this/get running with this? I know that compact metric spaces are totally bounded, so for any $\epsilon_{2}$ there's a finite collection of $\epsilon_{2}$-balls that cover $X$, but then we don't have any information about whether or not (any) of those balls are centered around our desired points. 
There's a little bit of information about $A$ that I don't think is needed - if it is, however, I'll gladly provide it. 
Thanks for any hints!
 A: Consider the open cover $B(a, \delta)$ for all $a \in A$ and all $\delta < \epsilon$. Since $X$ is compact, there is a finite subcover $B(a_i, \delta_i)$ for $i \in \{1, \ldots, n\}$. Now put $\epsilon_1 = \max(\delta_i : i \in \{1, \ldots, n\})$.
A: Let $x_1,...x_n$ be points such that $\bigcup_1^nB_\epsilon(x_i)=X$. For $x_1$ the function $d(x_1,y):X-\bigcup_2^n B_\epsilon(x_i)\to\Bbb R$ attains a maximum $m$ since the domain is compact. This $m$ must be smaller than $\epsilon$, so choose $m<\epsilon_1<\epsilon$. Then $B_{\epsilon_1}(x_1)\cup\bigcup_2^n B_{\epsilon}(x_i)$ still covers $X$, so we can replace the $\epsilon-$balls by $\mathcal U_1:=\{B_{\epsilon_1}(x_1),B_{\epsilon}(x_2),...,B_{\epsilon}(x_n)\}$. Again the set $X-\bigcup\mathcal U_1$ is compact, and by the same argument there is an $\epsilon_2<\epsilon$ such that $\mathcal U_2:=\{B_{\epsilon_1}(x_1),B_{\epsilon_2}(x_2),B_{\epsilon}(x_3),...,B_{\epsilon}(x_n)\}$ is still an open cover of $X$. Repeating this process gives you a finite number of values each smaller than $\epsilon$. Let $\delta=\max\{\epsilon_1,\epsilon_2,...,\epsilon_n\}$, then the balls $B_\delta(x_i)$ form an open cover of $X$.
We have proven that there exists a so-called shrinking of the original cover.
