Every sequence in $A$ has a Cauchy subsequence $\implies$ $A$ is totally bounded (Carothers, Theorem 7.5) 
Theorem 7.5: A set $A$ is totally bounded iff every sequence in $A$ has a Cauchy subsequence.

The $[\Rightarrow]$ direction of the proof is easy, I have trouble understanding the $[\Leftarrow]$ direction.
First, here are the definitions/characterizations of totally bounded I know:

$A\subset (M,d)$ is said to be totally bounded if for all $\epsilon>0$, there exist finitely many points $x_1,\ldots,x_n\in M$ such that $A\subset\bigcup_{i=1}^n B(x_i,\epsilon)$. Equivalently, $A\subset (M,d)$ is said to be totally bounded if for all $\epsilon>0$, there exist finitely many points $x_1,\ldots,x_n\in A$ such that $A\subset\bigcup_{i=1}^n B(x_i,\epsilon)$.
Also, $A$ is totally bounded iff for all $\epsilon>0$,  there are finitely many sets $A_1,A_2,...,A_n\subset A$ with $\text{diam}(A_i) < \epsilon$ for all $1\le i\le n$ such that $A\subset \bigcup_{i=1}^n A_i$.

I will reproduce the main parts of the proof of $[\Leftarrow]$ direction now.

Suppose $A$ is not totally bounded. Then there is some $\epsilon > 0$ such that $A$ cannot be covered by finitely many $\epsilon$-balls. Thus, by induction, we can find a sequence $(x_n)$ in $A$ such that $d(x_n,x_m) \ge \epsilon$ whenever $m\ne n$.

How? Where can I use induction? It is not obvious!

But then, $(x_n)$ has no Cauchy subsequence.

This makes sense! Once I have found an $(x_n)$ with the above properties, I know that it can't have a Cauchy subsequence. To see this, suppose it did - let $(x_{n_k})$ be the Cauchy subsequence of concern. Then we would want, $\forall \epsilon > 0, \exists N\in \Bbb N$ such that $p,q\ge N \implies d(x_{n_p},x_{n_q}) < \epsilon$. Sadly, we have $d(x_{n_p},x_{n_q}) \ge \epsilon$ for all $p\ne q$. Contradiction!
So, only the fact that such a $(x_n)$ exists bothers me - could I get some help in completing that part of the proof by induction or otherwise? Thanks!
 A: It's really by recursion, not induction (which is a proof technique).
So we have $\varepsilon >0$ such that $A$ cannot be covered by finitely many balls of radius $\varepsilon$.
Let $x_1 \in A$ to start the recursion. Then $A$ is not a subset of $B(x_1, \varepsilon)$ by assumption. So pick $x_2 \in A$ such that $x_2 \notin B(x_1, \varepsilon)$. Note that this means that $d(x_1,x_2) \ge \varepsilon$.
Now suppose we have already points $x_1, \ldots x_{m}$ so that $d(x_i, x_j) \ge \varepsilon$ for all $1\le i,j \le m, i \neq j$. By the assumption on $A$ we have
$$A \nsubseteq \bigcup_{i=1}^m B(x_i, \varepsilon)$$
or $A$ would have been coverable by finitely many $\varepsilon$-balls.
So pick $x_{m+1} \in A$ that is not in the union, and note that this means that $d(x_{m+1}, x_i) \ge \varepsilon$ for all $1 \le i \le m$.
So now we have $m+1$ many points $x_1,\ldots, x_m, x_{m+1}$ obeying $d(x_i, x_j) \ge \varepsilon$ for all $i \neq j$ so far. So we can do the recursion step going from $m$ to $m+1$ points, for every $m$.
So we have constructed a sequence $(x_n)_n$ by recursion, all of whose distances are $\ge \varepsilon$ between distinct indices.
The proof that it doesn't have a Cauchy subsequence is fine.
A: Suppose that $A$ is not totally bounded. Then, fix $\varepsilon>0$ such that, if $a_1,\ldots,a_n\in A$, then $A\not\subset\bigcup_{i=1}^nB(a_i,\varepsilon)$. Take $x_1\in A$. Since $A\not\subset B(x_1,\varepsilon)$, you can take $x_2\in A\setminus B(x_1,\varepsilon)$. Since $A\not\subset B(x_1,\varepsilon)\cup B(x_2,\varepsilon)$, there is some $x_3\in A\setminus B(x_1,\varepsilon)\cup B(x_2,\varepsilon)$. And so on… Then $m\ne m\implies d(x_m,x_n)\geqslant\varepsilon$.
A: Supposing, as in your quote, that $A$ is not totally bounded, and that $\varepsilon$ is a witness to this. Then we can construct a sequence as follows:

*

*Start with any point $x_0$ in $A$. We know that $A$ is not in a $\varepsilon$ ball around $x_0$ (else it would be totally bounded), so there is some point $x_1$ in $A \setminus B(x_0,\varepsilon)$.

*Repeat this process: after any finite number of steps $k$, we know that $A$ is not contained in $\bigcup_{i=0}^k B(x_i,\varepsilon)$, else it would be totally bounded, so we can choose $x_{k+1}$ to be any element of $A\setminus \bigcup_{i=0}^k B(x_i,\varepsilon)$, and we will have $d(x_{k+1},x_i) > \varepsilon$ for all $i \leq k$ by construction.

Since every point that we add is at least $\varepsilon$ from every previous point, all pairs of distinct points in our sequence are at least $\varepsilon$ apart from one another.
