Confused with proof that all Cauchy sequences of real numbers converge.

At that point in the proof you’re trying to show that if $n>N_2$, then $|a_n-L|<\epsilon$. If you can find a real number $x$ such that $|a_n-x|<\frac{\epsilon}2$ and $|x-L|<\frac{\epsilon}2$, the triangle inequality will give you the desired result, so the proof boils down to finding such an $x$.

1. What things do we know are close to $L$? Terms $a_{n_k}$ of the subsequence, provided that $k$ is sufficiently large.
2. What things do we know are close to $a_n$? Terms $a_m$ of the original sequence, provided that $m$ and $n$ are sufficiently large.

We take care of (1) first: there is an $N_1$ such that $|a_{n_k}-L|<\frac{\epsilon}2$ whenever $k>N_1$. This means that we can take our $x$ to be any $a_{n_k}$ with $k>N_1$, and we’ll have $|x-L|<\frac{\epsilon}2$.

Then we take care of (2): there is an $N_2$ such that $|a_n-a_m|<\frac{\epsilon}2$ whenever $m,n>N_2$. This means that since we’ve already specified that $n>N_2$, we can take our $x$ to be any $a_m$ with $m>N_2$, and we’ll have $|a_n-x|<\frac{\epsilon}2$.

Can we combine the two requirements? Is there an $a_{n_k}$ with $k>N_1$ that is also an $a_m$ with $m>N_2$? Equivalently, is there an $a_{n_k}$ with $k>N_1$ such that $n_k>N_2$? Sure: the sequence $\langle n_k:k\in\Bbb N\rangle$ is unbounded, so its tail $\langle n_k:k>N_1\rangle$ is also unbounded and contains a term $n_k>N_2$. Thus, we can set $x=a_{n_k}$ and satisfy both requirements, so that we have

$$|a_n-L|\le|a_n-x|+|x-L|<\frac{\epsilon}2+\frac{\epsilon}2=\epsilon$$

as desired. Note that this calculation doesn’t actually depend on a specific value of $k$: we could set $x=a_{n_\ell}$ for any $\ell\ge k$, the the calculation would be the same. As I said at the beginning, we’re really just trying to find one number $x$ that we can use to ‘tie’ $a_n$ to $L$ to within $\epsilon$; it turns out that there are lots of them, and it doesn’t matter which one we use.