Assume $X= (x_n)$ is not a Cauchy sequence. Prove that there exists $\epsilon>0,$ and a sub sequence $X'=(x_{n_k})$ such that $\left|x_{n_{k+1}}-x_{n_k}\right|\ge \epsilon, \forall k\in \mathbb N$.
My Attempt
Cauchy sequence: $\forall\epsilon>0\exists N\in\mathbb{N}:\forall (n,m\geq N\implies |x_m-x_n|<\epsilon) \tag{1}$
Negation: $\exists\epsilon>0\forall N\in\mathbb{N}:\exists n,m\geq N\wedge |x_m-x_n|\geq \epsilon\tag{2}$
Let $\epsilon$ be a given number
By(1), Stage I
For$N=1$, $\exists k_1,k_2\geq N\wedge |x_{k_1}-x_{k_2}|\geq \epsilon$
Choose $n_2=\max\{k_1,k_2\}$ and $n_1=\min\{k_1,k_2\}$
Stage 2
For$N=n_2$, $\exists k_3,k_4\geq N\wedge |x_{k_3}-x_{k_4}|\geq \epsilon$
Choose $n_4=\max\{k_3,k_4\}$ and $n_3=\min\{k_3,k_4\}$
$\vdots$
Stage m
For$N=n_{m}$, $\exists k_{m+1},k_{m+2}\geq N\wedge |x_{k_{m+2}}-x_{k_{m+1}}|\geq \epsilon$
Choose $n_{m+2}=\max\{k_{m+1},k_{m+2}\}$ and $n_3=\min\{k_{m+1},k_{m+2}\}.$
But when $k=2$, we can't conclude from this construction $\left|x_{n_{3}}-x_{n_2}\right|\ge \epsilon.$
How do I modify this construction to obtain the desired sequence $X'=(x_{n_k})$ such that $\left|x_{n_{k+1}}-x_{n_k}\right|\ge \epsilon, \forall k\in \mathbb N$?