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

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.

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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.

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.