I came across the following claim in this post.
Let $X$ be a metric space in which every infinite subset has a limit point. Then for every $\delta>0$ there exists $N_{\delta}\in \mathbb{N}$ and $x_i\in X, \, 1\le i\le N_{\delta}$ such that $$X \subseteq \bigcup_{i=1}^{N_\delta} B(x_i,\delta).$$
My attempt :
I shall try and do the problem via contraposition.
Suppose $\exists \delta>0$ such that $\forall n\in \mathbb{N}$ and for any finite subset $\{x_1,\dots x_n\}$ of $X$ we have $$X \nsubseteq \bigcup_{i=1}^{N_\delta} B(x_i,\delta).$$
Now, I need to exhibit an infinite subset of $X$ which has no limit point.
Let $p_1\in X.$ Now for each $k\in \mathbb{N}$ choose $p_{k+1}$ such that $$p_{k+1} \in X - \bigcup_{i=1}^{k} B(p_i,\delta).$$ This is possible since $X - \bigcup_{i=1}^{k} B(p_i,\delta)$ is non-empty.
Now, let $S = \{p_m : m\in \mathbb{N}\}.$ $S$ is an infinite subset of $X$ in which every element is atleast at a $\delta$ distance from all the other elements of $S.$ Hence for any $p\in X$, $B(p,\delta /2)$ can contain at most one point from $S$ and $p$ (therefore) cannot be a limit point.
Is my argument valid?Also please share if there are other approaches to the problem?
Yes, what you've done is pretty much the standard proof that a set which is not totally bounded contains a countably infinite set $\{p_n: n \in \mathbb{N}\}$ and $\delta>0$ such that $d(p_m, p_n) \ge \delta$ when $n \neq m$.
In the contrapositive form, this implies that every sequentially compact $X$ or limit point compact $X$, is totally bounded, the second of which is your original claim.