A metric space in which every infinite set has a limit point is separable I am struggling with one problem.
I need to show that if $X$ is a metric space in which every infinite subset has a limit point then $X$ is separable (has countable dense subset in other words).
I am trying to use the result I have proven prior to this problem, namely every separable metric space has a countable base (i.e. any open subset of the metric space can be expressed as a sub-collection of the countable collection of sets). 
I am not sure this is the right way, can anyone outline the proof?
Thanks a lot in advance!   

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.
