Question: 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!

Answer: 
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 $\epsilon\_1 = \max(\delta\_i : i \in \{1, \ldots, n\})$.
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](http://en.wikipedia.org/wiki/Shrinking_space) of the original cover.