Trouble with the proof of Cauchy convergence criterion

If $x_N − \frac{\epsilon}{3}<x_k <x_N + \frac{\epsilon}{3}$ holds for all $k>N$, then $ x_N-\frac{\epsilon}{3}≤ \inf_{k≥N} x_k$ and $ \sup_{k≥N} x_k ≤ x_N + \frac{\epsilon}{3}.$ Combine these two, to get the underbraced inequality and then the rest of the proof go through. Note, you hve used the generic $n$, which is not quite right. The inequalities follow one after the other, for $k>N$, which you fixed previously.

Maybe a more detailed proof will help with the ideas:

$(1).$ Since $(x_n)$ is Cauchy, it is bounded. And it's much easier to see this simply by taking $\epsilon=1$ and finding an integer $N$ such that $n,m>N\Rightarrow |x_n-x_m|<1$. Then, $|x_n|<\max\{|x_1|,\cdots |x_N|,|x_{N+1}|+1\}.$

$(2).$ Since $(x_n)$ is bounded, it has a convergent subsequence (I will give a proof of this at the end). So,

$(3).$ Let $x_{n_k}\to x$. Then, there is an integer $K$ such that if $k>K,\ |x-x_{n_k}|<\epsilon/2$. And there is an integer $N$ such that if $n>N, |x_m-x_n|<\epsilon/2.$ Now, choose an integer $l>K$ such that $n_l>N$. Then, if $n>N, |x_n-x|\le |x-x_{n_l}|+|x_n-x_{n_l}|<\epsilon.$

Proof of $(2):$ Let $(x_n)$ be a bounded sequence. Then there exists some $M > 0$ such that $|x_n| \le M$ for all integers $n$. Bisect the interval $[−M, M]$ into two closed intervals of equal length. One of these intervals must contain infinitely many $x_n.$ Let $I_1$ be that interval, and choose $x_{n_1}\in I_1$. Now, bisect $I_1$ into two closed intervals. Let $I_2$ be the subinterval of $I_1$ that contans infinitely many $x_n$, and choose one, $x_{n_2}\neq x_{n_1}$ such that $n_2>n_1$ (why is this possible?). In general, having constructed $\{I_j:I_j\subset I_{j-1},\ 1\le j\le k-1\}$, bisect $I_{k−1}$ into two closed intervals, one of which must contain infinitely many terms of $(x_n).$ Let $I_k$ be this closed interval, and choose $x_{n_k}\in I_k$ such that $n_k > n_{k−1}.$ Therefore, the induction proceeds and we obtain a subsequence $(x_{n_k})$ and a sequence of nested intervals $\{I_k\}_k$ whose diameters $|I_k|=M\cdot \left(\frac{1}{2}\right)^{k-1}$ tend to $0$ as $k\to \infty.$ This fact, and the Nested Interval Property, imply that the intersection $\bigcap I_k$ contains exactly one point $x$. Now let $\epsilon>0$ and choose $K$ such that $M\cdot \left(\frac{1}{2}\right)^{K-1}<\epsilon.$ Then, if $k>K,\ x_{n_k}\in I_k$ and $|I_k|<\epsilon.$ But, $x$ is contained in $\textit{every}\ I_k$ so $|x_{n_k}-x|<\epsilon.$ It follows that $x_{n_k}\to x.$