I study fixed point theory from Kirk and Khamsi's An Introduction to Metric Spaces and Fixed Point Theory and I couldn't understand a proof.
STEP 1: Let $\{d(x_n,x_{n+1})\}$ be a monotone decreasing sequence which is convergent to $0$.
Make use of this and additional conditions, it is shown with proof-by-contradiction that $\{x_n\}$ is a Cauchy sequence.
First, assume $\{x_n\}$ is not a Cauchy sequence. Then there exists $\varepsilon > 0$ such that for any $k \in \mathbb{N}$, there exist $m_k > n_k \geq k$, such that $$d(x_{m_k} ,x_{n_k} ) \geq \varepsilon.$$
Why are subsequences $\{x_{m_k}\} ,\{x_{n_k}\}$ used?
Furthermore, assume that for each $k$, $m_k$ is the smallest number greater than $n_k$ for which above inequality holds. In view of STEP 1 there exists $k_0$ such that $k > k_0 \Rightarrow d(x_k, x_{k+1}) \leq \varepsilon.$ For such $k$ we have
$$d(x_{{m_k}-1} ,x_{n_k}) < \varepsilon.$$
Why is $d(x_{{m_k}-1} ,x_{n_k}) < \varepsilon$ ?
"Why [are] subsequences $\{x_{m_k}\}$, $\{x_{n_k}\}$ used?"
Because that is how you negate the definition of a Cauchy sequence. That definition is that for large $n,m$ we have $d(x_m,x_n)<\varepsilon$. The negation is that you can't pick any integer, no matter how large, such that for larger $n,m$ that condition is met; i.e., for every integer $k$ there will always be $m_k> n_k$ such that $d(x_{m_k},x_{n_k})\ge\varepsilon$.
"Why [is] $d(x_{m_k-1},x_{n_k})\le\varepsilon$?"
Because we previously assumed that $m_k$ was the smallest number greater than $n_k$ for which the "above inequality," i.e. $d(x_{m_k},x_{n_k})\ge\varepsilon$, holds. Since $m_k-1$ is smaller than $m_k$, the latter inequality must be false if we replace $x_{m_k}$ by $x_{m_k-1}$, and so we must have $d(x_{m_k-1},x_{n_k})<\varepsilon$.