Question: Suppose for a metric space every infinite subset has a limit point. What should be my strategy to construct a countable dense subset there? Additionally, how do I intuitively guess that with such a property the metric space has a countable dense subset.

Answer: 
HINT: Let $\langle X,d\rangle$ be a metric space in which every infinite subset has a limit point. For each $n\in\Bbb N$ let $D\_n$ be a maximal subset of $X$ such that $d(x,y)\ge 2^{-n}$ whenever $x,y\in D\_n$ with $x\ne y$. (You can use Zorn’s lemma to show that $D\_n$ exists.)

* Show that each $D\_n$ is finite.
* Show that $\bigcup\_{n\in\Bbb N}D\_n$ is dense in $X$.

I’m not sure how you’d guess this result. The hypothesis on $X$ does tell you that $X$ does not contain an infinite closed discrete subset, which in some sense says that the points of $X$ aren’t spread out too much, but that property alone isn’t enough to ensure that $X$ is separable: the result really does use the fact that $X$ is a metric space as well.
In metric spaces, every infinite subset having a limit point (limited point compactness) is equivalent to compactness. 

Every compact metric space has a countable dense subset. See the answer to this question:

[Prove: Every compact metric space is separable](https://math.stackexchange.com/questions/190817/prove-every-compact-metric-space-is-separable)