So by definition of Cauchy, for all $\epsilon > 0$ and $i, j \in \mathbb{N}$, there exists an $M$ such that for all $i, j \geq M$, then $|x_i - x_j| < \epsilon'/2$ if we let $\epsilon = \epsilon'/2$.
By definition of convergence of a subsequence, for all $\epsilon'' > 0$ and $n_k \in \mathbb{N}$, there exists an $N$ such that for all $n_k \geq N$, then $|x_{n_k} - x| < \epsilon'''/2$ if we let $\epsilon'' = \epsilon'''/2$.
I know the triangle inequality comes into play later, so that for all $\epsilon'''' > 0$ and $n \in \mathbb{N}$, there exists an $L$ such that for all $n \geq L$, $|x_n - x_{n_k}| + |x_{n_k} - x| < \epsilon''''$.
I understand that in the above inequality that we want, we already have that $|x_{n_k} - x| < \epsilon'''/2$, so all we need to do to complete the proof is somehow make $|x_n - x_{n_k}| < \epsilon'/2$. I've read a lot of proofs of this, but they don't explain why they are doing what they are doing and make huge logical jumps as obvious statements. I also don't know why $\epsilon'/2 + \epsilon'''/2 = \epsilon''''$.