Q:

will a nonempty countable and compact subset of a metric space always contain an isolated point?

Let $(X,d)$ be a metric space, $K \subseteq X$ is nonempty, countable, and compact.
I could not come up with an example where such a K has no isolated point(s).
so I want to prove that an isolated point will always exist.
I tried by first assuming that no point is isolated, then I want to come up with an open cover which will not admit a finite subcover. I don't know how to proceed next.

A:

A subset of a topological space is said to be perfect if it is closed and has no isolated points. It can be proved (see, for example, Proof that a perfect set is uncountable) that a perfect set is uncountable
So in fact $K$ must have isolated points. Let $K$ be a compact, countable subset of the metric space $(X,d)$ and assume that it has no isolated points. Since $K$ is compact, it is closed in $X$, and so it must be perfect. But then, by the above, $K$ is uncountable, contradicting the assumption that $K$ was countable.
—
Edit: @DanielFischer pointed out that completeness is necessary for my statement above! As noted in the comments, a perfect subset of a non complete space can be countable. 

A:

Assume that $K$ is countable, compact, non-empty and that has no isolated points. Say that $K=\{x_1,x_2,\ldots\}$. (These names are yet to be assigned).
Take $x_1\in K$, and an open set $U_1\subset X$ such that $x_1\in U_1$.
For $n\ge 2$, since $K$ has no isolated points, there is a point $x_n\in U_{n-1}\setminus\{x_{n-1}\}$. Let $U_n$ be an open set such that $x_n\in U_n$ and $\overline{U_n}\subset U_{n-1}\setminus\{x_{n-1}\}$ (this open set exists because $K$ is metric). Define $F_n=\overline{U_n}\cap K$.
Note that each $F_n$ is compact, because is a closed contained in a compact.
But $F=\bigcap_{n=1}^\infty F_n$ is not empty, and $x_n\notin F$ for any $n\ge 1$.
This proves that no sequence in $K$ covers $K$. Contradiction.

