Metric Spaces (Extreme Value Theorem)

Question

I been reading Royden and Fitzpatrick's Real Analysis and in chapter 9 - 4th edition ($*$) they prove the extreme value theorem for metric spaces.

First I'll transcribe the theorem and then I will state my question, which concerns one of the steps in the proof (Any necessary definitions will be at the bottom):

Theorem 22 (Extreme Value Theorem): Let $X$ be a metric space. Then $X$ is compact if and only if every continuous real valued function on $X$ takes a maximum and a minimum value.

What interests me is the if part, so i'll cut to the chase. Their proof is as follows:

"To prove the converse, assume every continuous real valued function on $X$ takes a maximum and a minimum value. According to Theorem 20, to show that $X$ is compact it is necessary and sufficient to show it is totally bounded and complete. We argue by contradiction to show that $X$ is totally bounded. If $X$ is not totally bounded, then there is an $r > 0$ and a countably infinite subset of $X$, which we enumerate as $\{x_n\}_{n = 1}^{\infty}$, for which the collection of open balls $\{B(x_n;r)\}_{n = 1}^{\infty}$ is disjoint.

The sentence in bold is the heart of the problem. From earlier proofs, given the definition of a sequentially bounded set, it was shown that, for some $r > 0$, it was possible to inductively create a sequence $\{x_n\}_{n = 1}^{\infty}$ for which \begin{equation} \forall \; n \in \mathbb{N} \; \; \; x_n \notin B(x_i;r), \; \; \forall \; i = 1, ..., n - 1, \end{equation} but at no point was it stated that the open balls had to be disjoint and, from what I can see, they don't have to be. So my question is: how do I prove the statement in bold?

Definition: A metric space $X$ is said to be totally bounded provided for each $\epsilon > 0$, the space $X$ can be covered by a finite number of open balls of radius $\epsilon$.

($*$) The statement and proof can be found in "Real Analysis" (4th edition), pages 200-201.

Answer 1

I think the argument of choosing a sequence $\{ x_n \}$, that is your paragraph,

... for some $r>0$, it was possible to inductively create a sequence $\{ x_n \}_{n=1} ^\infty$ for which $$ \forall n \in \mathbb{N}\, , x_n \not \in B(x_i, r)\, , \quad \forall i = 1, \dots, n-1 $$

still works, except that after $\{ x_n \}$ has been chosen, we instead consider the collection of open balls $\{ B(x_n, \frac{r}{3})\}$. This collection will be disjoint, since any two centers are at least $r$ unit of distance apart.