TITLE: will a nonempty countable and compact subset of a metric space always contain an isolated point?
QUESTION [8 upvotes]: 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.

REPLY [1 votes]: We assume throughout this answer that

$(X,d)$ is a nonempty metric space containing no isolated points.

Proposition 1: If $x \in X$ a closed ball $B$ can be found disjoint from $\{x\}$.
Proof
$X$ must contain a point $y$ not equal to $x$. Setting $\alpha = d(x,y)$, the closed ball $B_y$ with center $y$ and radius $\frac{\alpha}{2}$ can't contain $x$. $\qquad \blacksquare$
Proposition 2: If $B$ is any closed ball in $X$ and $b \in B$ then there exists a closed ball $B^{'} \subset B$  that doesn't contain the point $b$.
Proof
If $b$ is not the center $b_0$ of $B$, simply use the closed ball about $b_0$ of radius $\frac{d(b_0,b)}{2}$.
If $b = b_0$, let $\alpha_0$ be the radius $B$. Choose a point $b^{'}$ in the interior of $B$ not equal to $b_0$ and set $\alpha = d(b_0,b^{'})$. Setting $\alpha^{'}  = \frac{1}{2} \text{min(}\alpha_0 - \alpha\text{, }\alpha\text{)}$, it is elementary checking that the closed ball $B^{'}$ about $b^{'}$ of radius $\alpha^{'}$ is contained in $B$ and doesn't contain $b$. $\qquad \blacksquare$
Proposition 3: If $X$ is a compact space then a countable listing $(x_n)_{\,n \ge 0}$ with $x_n \in X$ can never be a complete enumeration for $X$.
Proof
By applying proposition 2 (use proposition 1 for the initial closed ball, $B_0$)
construct a non-increasing chain of closed balls satisfying
$\tag 1 X \supset B_0 \supset B_1 \supset B_2 \dots \supset B_k \supset \dots $
$\tag 2 x_j \notin B_k \; \text{ for } \, 0 \le j \le k$
Since $X$ is compact, we can apply Cantor's intersection theorem. So there exists a point $\hat z$ belonging to all the $B_k$, and by design that point can't be in the $x_n$ listing. $\qquad \blacksquare$
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