Lemma: Let $(X,d)$ be a metric space. For each $n \in \{1,2,3,\ldots\}$ there is (by Zorn's lemma) a maximal (by inclusion) set $D_n$ such that for all $x,y \in D_n$ with $x \neq y$, we have $d(x,y) \ge \frac{1}{n}$. If all $D_n$ are at most countable, then $X$ is separable.
Proof: We'll show that $D = \cup_n D_n$ is dense (and is countable when all $D_n$ are), and so pick any $x \in X$ and any $r>0$ and we'll show that $B(x,r) \cap D \neq \emptyset$. Suppose not, then find $m$ with $\frac{1}{m} < r$. Then from $B(x,r) \cap D = \emptyset$ we know that
$d(x, y) \ge \frac{1}{m}$ for all $y \in D_m$ (or else $y \in D \cap B(x,r)$), and then $D_m \cup \{x\}$ would contradict the maximality of $D_m$. So the intersection with $D_m$ and hence $D$ is non-empty, and so $D$ is dense.
(The above is the heart of the proof that a ccc metric space is separable, as $X$ ccc implies that all $D_m$ are countable, as the $B(x,\frac{1}{m})$, $x \in D_m$ are a pairwise disjoint open family of non-empty sets.)
Now suppose that $(X,d)$ is countably compact. Suppose for a contradiction that $X$ is not separable. Then for some $m \ge 1$ we have that $D_m$ (as in the lemma) is uncountable.
Fix such an $m$.
Now $D_m$ is discrete (clear as $B(x,\frac{1}{m}) \cap D_m = \{x\}$ for each $x \in D_m$) and closed: suppose that $y \in X\setminus D_m$ is in $\overline{D_m}$, then $B(y, \frac{1}{2m})$ contains infinitely many points of $D_m$ and for any $2$ of them, say $x_1, x_2 \in D_m$, we'd have $d(x_1, x_2) \le d(x_1, y) + d(y,x_2) < \frac{1}{m}$ contradiction (as points in $D_m$ are at least $\frac{1}{m}$ apart). So $D_m$ is closed and discrete (as an aside: being uncountable this would already contradict Lindelöfness of $X$; this shows that a Lindelöf metric space is separable, e.g.), but we want to contradict countable compactness, which is easy too:
Choose $A \subseteq D_m$ countably infinite (so that $A$ is closed in $X$, as all subsets of $D_m$ are), and define a countable open cover $\mathcal{U} = \{B(p, \frac{1}{m}): p \in A\} \cup \{X\setminus A\}$ of $X$ that has no finite subcover: we need every $B(p,\frac{1}{m})$ to cover $p$ for all $p \in A$.
This contradiction then shows $X$ is separable (and thus has a countable base ,is Lindelöf etc. finishing the compactness).
The crucial fact is that all of the following are equivalent for a metric space:
- $X$ has a countable base.
- $X$ is separable.
- All discrete subspaces of $X$ are at most countable.
- All closed and discrete subspaces of $X$ are at most countable.
- $X$ is ccc.
- $X$ is Lindelöf.
In the above I essentially did "not (2) implies not (4)" implicitly. The fun is that countable compactness implies (4) easily (such closed discrete subspaces are even finite) and thus we get Lindelöfness "for free". Also implicit in the above proof is that every countably compact space is limit point compact.
Best Answer
So, I think I have an idea and would like to run it past someone to see if it holds. I have already proved that if $(X,d)$ is metric and limit point compact then:
Proof: Let $A$ be an arbitrary indexing set and $\{U_\alpha\}_{\alpha\in A}$ be an open cover of $X$. Now, $\exists\delta>0$ so that if $D\subseteq X$ such that diam($D$)$<\delta$, then $\exists U_{\alpha_D}$ such that $D\subseteq U_\alpha$. Also, $\exists \{x_1,\dotsc,x_k\}\subseteq X$ such that $X=\bigcup_{i=1}^k B_{\delta/3}(x_i)$. Both of the facts follow from the two statements above.
Now, $\forall 1\leq i\leq k$, since diam($B_{\delta/3}(x_i)$)$<\delta$, $\exists U_{\alpha_i}$ such that $B_{\delta/3}\subseteq U_{\alpha_i}$. So if $x\in X$, then $\exists1\leq n\leq k$ such that $x\in B_{\delta/3}(x_n)\subseteq U_{\alpha_n}$.
So $\{U_{\alpha_i}\}_{i=1}^k$ is a finite subcover of $X$. Therefore $(X,d)$ is compact.