Countable choice implies that a metric space is limit point compact if and only if it is compact.
From there it is not hard to deduce separability, as in ZFC.
$\newcommand{AC}{\mathsf{AC}}$
Let me point out some key points in a way for a proof of the above.
First we observe that under $\AC_\omega$ every infinite set has a countable subset. Let us see that it also implies that "every infinite set has an accumulation point" implies that every infinite sequence has a convergent subsequence:
Suppose that $\{x_n\mid n\in\omega\}$ is an infinite sequence. It has an accumulation point by the assumption. Namely a point $x$ such that for every open neighborhood $U$, there is some $x_n\neq x$ such that $x_n\in U$. Since we are in a metric space, without loss of generality this means that for every $k$ there is some $x_n\in B(x,\frac1k)$ and $x\neq x_n$.
Assume that $x\neq x_n$ for all $n$, just for simplicity, now it is clear that in every $B(x,\frac1k)$ there are infinitely many points from the sequence, otherwise there were only $m$ many points in $B(x,\frac1k)$ for some $k$, and we could then take $k'>k$ to be large enough so that none of these points appear in $B(x,\frac1{k'})$.
Since we have infinitely many, let $x_{n_k}$ be such that $n_k = \min\{n\mid x_n\in B(x,\frac1k)\}$. Note that if $k>j$ then $n_k\geq n_j$ since $x_{n_k}$ is closer to $x$ than $x_{n_j}$. If there are repetitions we can remove them by induction. We don't use the axiom of choice at all here, since this is already a well-ordered set.
Now it is not hard to see that $x_{n_k}\to x$, since for every $\varepsilon>0$ there exists $K$ such that for $k>K$ we have $x_{n_k}\in B(x,\varepsilon)$.
Conclusion I: $X$ is sequentially compact. Now let us show that it is compact.
The implication now is as in Brian's answer.