Theorem 2.6.4 in Understanding Analysis by Abbott states,
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
A sequence converges if and only if it is a Cauchy sequence.
$(\rightarrow)$ direction.
This direction has already been established earlier.
$(\leftarrow)$ direction.
Strictly speaking, wouldn't it be the case that,
(1) $\exists N_1$, such that the distance $\lvert x_n - x_m \rvert$ can be made smaller than $\epsilon/2$ for all $m,n \ge N_1$. That is,
\begin{align*} \lvert {x_n - x_m} \rvert < \frac{\epsilon}{2} \end{align*}
for all $m,n \ge N_1$
(2) $\exists N_2$, such that the distance $\lvert x_{n_k} - x\rvert$ can be made smaller than $\epsilon/2$ for all $n_{k} \ge N_2$:
\begin{align*} \lvert {x_{n_k} - x} \rvert < \frac{\epsilon}{2} \end{align*}
Choose a fixed term $x_{n_K}$ in the subsequence $(x_{n_k})$, so that $n_K \ge N_2$.
Let $N = \max \{N_1,N_2\}$. To see that $N$ has the desired property (for the original sequence $(x_n)$), observe that if $n \ge N$, then
\begin{align*} \absval{x_n - x} &= \absval{x_n - x_{n_K} + x_{n_K} - x}\\ &\le \absval{x_n - x_{n_K}} + \absval{x_{n_K} - x}\\ &< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align*}
Is the minor detail of choosing $N = \max \{N_1,N_2\}$ ommitted and implicit in the textbook proof? Or this isn't required.