See the proof below about why a complete metric space with no isolated points must be uncountable.
I don't quite understand why $y\neq x_n$ for any $n$. I think it might be possible to show this if I can show $y_n\neq x_i$ for any $1\leq i\leq n,$ but do to this, I found it easier to use the additional condition that $x_i\not\in \overline{B}_{r_n}(y_n)$ for any $1\leq i\leq n.$
To clarify, I need to satisfy $y_n \in \overline{B}_{r_{n-1}}(y_{n-1}),y_n\neq x_n$ and $\overline{B}_{r_n}(y_n)\subseteq B_{r_{n-1}}(y_{n-1})$ and $0<r_n < \frac{1}n$ and $x_n\not\in \overline{B}_{r_n}(y_n)$.
And finally, aren't all the open balls of $X$ infinite?
First observe that each open ball $B_r(z)$ must contain more than one point: If $B_r(z)$ were a singleton, then $B_r(z) =\{z\}$ so that $z$ would be an isolated point. You can use this to show that all open balls $B_r(z)$ are infinite, but that is irrelevant for the argument.
Your book recursively constructs $y_n \in X$ and $r_n \in (0,\frac 1 n)$ such that
$y_n \ne x_n$
$\bar B_{r_{n+1}}(y_{n+1}) \subset B_{r_n}(y_n)$
but misses to say that we also want to have
The third condition is mentioned only for $n = 1$.
The recursive construction is based on the above-mentioned fact: We can always choose $y_{n+1} \in B_{r_n}(y_n)$ such that $y_{n+1} \ne x_{n+1}$. Given $y_{n+1}$, we can choose $r_{n+1} \in (0, \frac 1 {n+1})$ such that 2. and 3. are satisfied.
So why is $y \ne x_n$ for all $n$? Assume $y = x_n$ for some $n$. Then $y \notin \bar B_{r_n}(y_n)$ by 3. This is a contradiction because $y_k \in \bar B_{r_n}(y_n)$ for $k \ge n$ which implies $y \in \bar B_{r_n}(y_n)$.