Show that a compact metric space $X$ is locally compact Assume that $X$ is a compact metric space, that is by definition, every sequence in $X$ has a convergent subsequence.
Locally compact means that every point in $X$ has a compact neighbourhood. That is, for every point in $x \in X$ we can find an open ball for which every sequence has a convergent subsequence. 
So now what I need to do is take an arbitrary sequence $(x_{n})$ in a ball around $x_0 \in X$ and show that it has a convergent subsequence. How do I do that? 
 A: As you just want some compact neighbourhood of $x \in X$ you can just take the whole space $X$ itself. Notice that since compact metric spaces are bounded this is also an open ball centered at $x$.
Often this is the only choice for a compact neighbourhood of $x$ which is also an open ball: As every compact subset of a metric space is closed (because metric spaces are Hausdorff) every such ball must be clopen. So if $X$ is connected, for example $X = [0,1]$, we have no other choice.

I think it is important to point out that your definition of a locally compactness does not require the compact neighbourhood of $x$ to be an open ball, which you (incorrectly) seem to assume.
Not only makes this assumption the given problem more difficult, but it can fail you for non-compact spaces: Take for example the real line $\mathbb{R}$ with the standard euclidian metric. This is a locally compact metric space (for every $x \in \mathbb{R}$ the set $[x-1,x+1]$ is a compact neighbourhood of $x$), but as $\mathbb{R}$ is connected we find by the above argumentation that a compact, open ball centered at $x \in \mathbb{R}$ would need to be the whole space, which is absurd.

I would also like to point out that a compact metric space $X$ does already satisfy the alternative and stronger definition of a locally compact space, namely that every $x \in X$ has a neighbourhood basis of compact neighbourhoods, i.e. that every neighbourhood $U$ of $x$ contains a compact neighbourhood. To see this notice that there exists some $\varepsilon > 0$ with $B_{2\varepsilon}(x) \subseteq U$, which is why $\overline{B_{\varepsilon}(x)} \subseteq U$. This is a neighbourhood of $x$, as it contains $B_{\varepsilon}(x)$, and as a closed subset of the compact space $X$ it is compact itself.
A: You can let the open ball be any open ball, and since $X$ is compact metric, every sequence in that open ball has a convergent subsequence. This is pretty much trivial.
