My attempt:
Let $X$ be a metric space in which every infinite set has a limit point .We fix $\delta >0$ and we pick $x_1 \in X$ .Having chosen $x_1,...,x_j \in X$., we choose $x_j+1 \in X$ so that $d(x_i,x_j) \ge \delta$ where $i=1,2,3,...,j$.This process must stop after a finite number .If not then we can find a sequence $(x_n)$ such that it has no limit point ,which is a contradiction to our assumption.
We pick $\delta =1/n$($n=1,2,...$).Then we can find $(x_1(1/n),...,x_k(1/n))$ for each $1/n$ such that the above condition holds true .Now, $A=\cup B_d(x_i(1/n),1/n)$ will cover $X$. We try to show that this is a countable dense subset , let $x \in X$ be a point then we take the neighborhoods $B_d(x,r)$ where $r>0$ .Now , we choose $1/N<r$ and we see whether $B_d(x,r)$ contains any point of $x_1(1/n),...,x_k(1/n)$ for $n \ge N$.If not then we can conclude that $d(x_i,x) \ge 1/N$ then $x \in A$.
So this has been my attempt.I am sorry if my notations are confusing but I have tried my best to make it comprehensive. Can someone point out my mistake in the proof or how should I continue from here.
Your idea is good. What you have to do get a complete proof is to choose a maximal set $(x_1(\frac 1 n).,..., x_k(\frac 1 n))$ for each $n$. In other words choose $k$ to be the largest integer such that a finite set with $k$ elements has points separated by distance $\frac 1 n$. Then your last step will lead to a contradiction, finishing the proof.