Prop: Every sequence has a Cauchy subsequence. $(X,d)$ is a totally bounded metric space. 
Prop: Every sequence has a Cauchy subsequence.
Since $(X,d)$ is totally bounded metric space , for every $r>0$, there exist finitely many points $x_1,...,x_n \in X$ such that $X \subseteq \bigcup^n_{i=1}B_{r}(x_i)$. Assume that $\{p_n\}_{n\in \mathbb{N}}$ is a sequence in $X$. Then, one of these balls must contain $\{p_n\}_{n\in \mathbb{N}}$ for infinitely many $n$, which gives us a subsequence $\{p_{n_k}\}_{k \in \mathbb{N}}$ with $d(p_{n_l}, p_{n_m}) \leq d(p_{n_l},x_i) + d(x_i, p_{n_m}) < \frac{r}{2} + \frac{r}{2}=r$
for all $l, m \in \mathbb{N}$. At this point, is it possible to use induction to conclude that the subsequence is Cauchy without depending on the choice of radius?
 A: You need to do a recursive construction that involves a sequence of radii converging to $0$. I’ll use radii $2^{-n}$ for $n\in\Bbb N$.
Start with your sequence $\langle p_n:n\in\Bbb N\rangle$. $X$ has a finite cover by open balls of radius $1$, so there is an infinite $A_1\subseteq\Bbb N$ such that for all $k,\ell\in A_1$, $d(p_k,p_\ell)<2\cdot 1=2$. Similarly, $X$ has a finite cover by open balls of radius $\frac12$, and $A_1$ is infinite, so there is an infinite $A_2\subseteq A_1$ such that for all $k,\ell\in A_2$, $d(p_k,p_\ell)<2\cdot\frac12=1$. 
In general, if $A_n$ is an infinite subset of $\Bbb N$, $X$ has a finite cover by open balls of radius $2^{-n}$, so there is an infinite $A_{n+1}\subseteq A_n$ such that for all $k,\ell\in A_{n+1}$, $d(p_k,p_\ell)<2\cdot2^{-n}=2^{-n+1}$.
Now let $k_1=\min A_1$, and for $\ell\in\Bbb N$ let $k_{\ell+1}=\min(A_\ell\setminus\{k_1,\ldots,k_\ell\})$. Show that $\langle p_{k_i}:i\in\Bbb Z^+\}$ is a Cauchy subsequence of $\langle p_n:n\in\Bbb N\rangle$.
