TITLE: How to show that$(X,d)$ is totally bounded
QUESTION [5 upvotes]: How to show that a metric space $(X,d)$ is totally bounded $\iff$ every infinite subsets of $X$ contains distinct points which distinct points that arbitrarily close to each other.
I don't know how to prove $\Leftarrow.$ Please help. 
BTW I'm looking for a little hint !

REPLY [2 votes]: Suppose that $\langle X,d\rangle$ is not totally bounded. Then there is an $\epsilon>0$ such that for all finite $F\subseteq X$, $\{B(x,\epsilon):x\in F\}$ does not cover $X$, i.e., $X\setminus\bigcup_{x\in F}B(x,\epsilon)\ne\varnothing$. This implies that we can recursively construct a set $A=\{x_n:n\in\Bbb N\}\subseteq X$ such that 
$$x_{n+1}\in X\setminus\bigcup_{k\le n}B(x_k,\epsilon)$$
for each $n\in\Bbb N$. $A$ is an infinite subset of $X$; does it contain distinct points that are arbitrarily close to each other?