A metric space in which every infinite set has a limit point is separable

Let $\langle X,d\rangle$ be a metric space in which each infinite subset has a limit point. For any $\epsilon>0$ an $\epsilon$-mesh in $X$ is a set $M\subseteq X$ such that $d(x,y)\ge\epsilon$ whenever $x$ and $y$ are distinct points of $M$. Every $\epsilon$-mesh in $X$ is finite, since an infinite $\epsilon$-mesh would be an infinite set with no limit point. Let $\mathscr{M}(\epsilon)$ be the family of all $\epsilon$-meshes in $X$, and consider the partial order $\langle \mathscr{M}(\epsilon),\subseteq\rangle$. This partial order must have a maximal element: if it did not have one, there would be an infinite ascending chain of $\epsilon$-meshes $M_0\subsetneq M_1\subsetneq M_2\subsetneq\dots$, and $\bigcup_n M_n$ would then be an infinite $\epsilon$-mesh. Let $M_\epsilon$ be a maximal $\epsilon$-mesh; I claim that $$X=\bigcup_{x\in M_\epsilon}B(x,\epsilon)\;,$$ where as usual $B(x,\epsilon)$ is the open ball of radius $\epsilon$ centred at $x$. That is, each point of $X$ is within $\epsilon$ of some point of $M_\epsilon$. To see this, suppose that $y\in X\setminus \bigcup\limits_{x\in M_\epsilon}B(x,\epsilon)$. Then $d(y,x)\ge\epsilon$ for every $x\in M_\epsilon$, and $M_\epsilon \cup \{y\}$ is therefore an $\epsilon$-mesh strictly containing $M_\epsilon$, contradicting the maximality of $M_\epsilon$.

Now for each $n\in\mathbb{N}$ let $M_n$ be a maximal $2^{-n}$-mesh, and let $$D=\bigcup_{n\in\mathbb{N}}M_n\;.$$ Each $M_n$ is finite, so $D$ is countable, and you should have no trouble showing that $D$ is dense in $X$.

A proof based on Rudin's Hints (Page 45, Qn 24)

Step 1: Fix $ \delta >0$, and pick $x_{1}\in X$. Having chosen $x_{1},...,x_{j}\in X$, choose $x_{j+1}\in X$, if possible, so that $d(x_{i},x_{j+1})\geq \delta $ for $i =1,...,j.$ This process must stop after a finite number of steps, otherwise $x_{i}\in X, i\in N$ is an infinite set in X, so it should have a limit point, say $x\in X$. Then any neighborhood of $x$ with radius less than $\frac{\delta}{2}$ contain at most one term of the sequence (remember, any two distinct terms of the sequence are of atleast $\delta$ distance). A contradiction.

Thus X can be covered by finitely many neighborhoods of radius $\delta$.

Step 2: Take $\delta = \frac{1}{n}$ ($n = 1,2,3,...$). Let $\{x_{n_{1}},...x_{n_{k(n)}}\}$ be the finite set obtained from step 1 corresponding to $\delta=\frac{1}{n}$. Let $D = \cup_{n=1}^{\infty} \{x_{n_{1}},...x_{n_{k(n)}}\}$. Then D is countable.\Next we prove D is dense in X which will prove the result.

If $D=X$ nothing to prove, otherwise let $x\in X \setminus D$ and take an $\epsilon$ - neighborhood of $x$. Choose n such that $\frac{1}{n}<\epsilon$. Neighborhoods of $x_{n_{1}},...x_{n_{k(n)}}$ with radius $\frac{1}{n}$ will cover X. So $x$ will be in one of such neighborhoods, say neighborhood of $x_{n_{i}}$, hence $d(x,x_{n_{i}})<\frac{1}{n}<\epsilon$. Thus $x$ is a limit point of D.

Since $X$ is limit point compact, it is totally bounded. That is, for every $\epsilon>0$, there is a finite cover of $X$ consisting of balls of radius $\epsilon$ (if not, one could construct a sequence in $X$ that has no limit point). For each positive integer $n$, let $A_n$ a finite cover of $X$ of open sets of radius $1/n$. Now consider $\cup C_n$, where $C_n$ is the set of centers of the elements in $A_n$.

This is essentially what Brian did, but I'll post it anyway. In a nutshell, you are showing: limit point compactness implies $X$ is totally bounded and a totally bounded space is seperable.