Question About Compactness

Question

I want to prove the following- assume $X$ is a compact metric space. For all $\epsilon > 0$, there exists a natural number $n$ and $x_1....x_n \in X$ such that $X$ is equal to the finite union of open balls $B(x_i, \epsilon)$. I know that I need a particular finite open cover depending on epsilon to satisfy the condition but Im kind of stuck.

Answer 1

If you are not familiar with the definition of compactness, in this case it means that every open cover of $X$ has a finite subcover. As G. Sassatelli commented, take the open cover $\{B(x,\epsilon): x \in X\}$. The proof should be straightforward from this point.