Every subsequence of a Cauchy sequence is a Cauchy sequence

A book on topology that I am reading tells me to

Prove that every subsequence of a Cauchy sequence is a Cauchy sequence.

which I do not believe is true. Here is my counterexample:
The sequence $$\{x_n\}$$ with $$x_n=\frac{1}{n}$$ for positive integer $$n$$ is Cauchy (because it converges to $$0$$, and convergence implies Cauchy). Its subsequence $$1,\frac{1}{2},\frac{1}{3}$$ is not Cauchy because the distance (with the Euclidean metric) is at least $$\frac{1}{6}$$ which is not an arbitrarily small $$\epsilon$$. Is this counterexample valid?

Here are the definitions:

1. A sequence $$\{x_n\}$$ of points in a metric space $$(X,d)$$ is said to be a Cauchy sequence if given any real number $$\epsilon>0$$, there exists a positive integer $$n_0$$ such that for all integers $$m\geq n_0$$ and $$n\geq n_0$$, $$d(x_m,x_n) \leq \epsilon$$.
2. If $$\{x_n\}$$ is any sequence, then the sequence $$x_{n_1}, x_{n_2}, ...$$ is said to be a subsequence if $$n_1
• Subsequences are supposed to have infinitely many terms. That is what $'...'$ in the definition indicates. Nov 17, 2022 at 9:31
• So $x_n=1/6$ for $n>3$ and this is enough... Nov 17, 2022 at 9:54

As geetha290krm pointed out in the comments section, the issue with your counterexample is that a subsequence needs to itself be a sequence, and thus have infinitely many terms. With this in mind, we can prove the given statement. Specifically the thing to notice is that if $$\{x_{n_i}\}_{i \in \mathbb{N}}$$ is a subsequence of a sequence $$\{x_n\}_{n \in \mathbb{N}}$$, then we must have $$n_i \geq i$$ for all $$i \in \mathbb{N}$$. So then given $$\epsilon > 0$$, we know that there is some $$N_\epsilon$$ where $$i,j \geq N_\epsilon$$ implies $$|x_i - x_j| < \epsilon$$ since the original sequence is Cauchy. Since $$n_i \geq i$$ and $$n_j \geq j$$, we have that $$n_i, n_j \geq N_\epsilon$$, so then $$|x_{n_i} - x_{n_j}| < \epsilon$$, and we see that the subsequence is Cauchy as well.