Prove that a metric space is totally bounded if and only if every sequence has a Cauchy subsequence.
I think I proved the Cauchy subsequence part:
Let $a_{0},a_{1}, a_{2}, a_{3}, a_{4},…\in X$ be a sequence.
For each $k$, let $F \subseteq X$ be a finite $\frac1k$-net.
Given $I \subseteq \Bbb N_{\ge0}$ and $k>1$ you find an infinite $J \subseteq I$ such that:
$$\exists p\in F:\forall n\in J: d(x_n,p)<\frac1k$$
Best Answer
You have proved that totally bounded implies every sequence has Cauchy subsequence, so I will prove the other implication. This is a proof using the contrapositive, that is, not totally bounded implies that there is a sequence with no Cauchy subsequence.
Suppose that $X$ is not totally bounded. Then there is an $\epsilon>0$ such that for all finite sets of points $\{x_1,\ldots,x_n\}$ $$X\neq \bigcup_{k=1}^n B(x_k;\epsilon).$$ Now we construct a sequence that has no Cauchy subsequence. Start with a finite collection of points $\{x_1,\ldots,x_n\}$, as above. Then since $X\neq \bigcup_{k=1}^n B(x_k;\epsilon)$, there is a point $x_{n+1}\in X$ such that $x_{n+1}\notin \bigcup_{k=1}^n B(x_k;\epsilon)$. Moveover, $$X\neq \bigcup_{k=1}^{n+1} B(x_k;\epsilon)$$ because if it were equal, we would have a contradiction to our assumption. Wash, rinse, repeat this process to get a sequence $(x_k)_{k=1}^\infty$.
To check that this has no Cauchy subsequence, notice that for any two terms $x_n$ and $x_m$ in this sequence, if $m>n$ then $$x_m\notin \bigcup_{k=1}^{m-1} B(x_k;\epsilon).$$ In particular, $x_m\notin B(x_n;\epsilon)$, hence $d(x_m,x_n)\geq \epsilon$. Similarly if $n>m$. This shows that the terms of this sequence are at least $\epsilon$ in distance apart, hence no Cauchy subsequence can exist.