This is a problem from my metric space book,I want to know if my proof is correct and whether the proof can be simplified:
Statement: Suppose $X$ is a totally bounded and $A\subset X$ is infinite,then for each $\epsilon>0$,$A$ has an infinite subset $B$ such that $diam(B)\leq \epsilon$.
Proof: Since $A$ is infinite,we can take a sequence of distinct terms from $A$,Now this sequence $(x_n)$ must have a Cauchy subsequence by the total boundedness of $X$(sequential criterion).Suppose the subsequence is $(x_{r_n})$.Then given $\epsilon>0$,we can find $n_\epsilon\in \mathbb N$ such that $\forall m,n\geq n_\epsilon$,$d(x_{r_n},x_{r_m})<\epsilon/2$.So after a given stage $n_\epsilon$,all the terms fit in a ball of radius $\epsilon/2$ centred at $x_{r_{n_\epsilon}}$.So if we take $B=\{x_{r_n}:n\geq n_{\epsilon}\}$,which is infinite as terms are distinct ,then $diam(B)\leq \epsilon$.
Is the above proof alright?