This is an attempt of a proof of a rather basic result.
Proposition: The discrete metric $d$ is complete in $\mathbb{N}$.
Proof:
Let $x_n$ be an arbitrary sequence in $\mathbb{N}$ endowed with its natural discrete metric, and assume it to be Cauchy. Thus, by the definition of Cauchy sequence in a generic metric space $(X, \rho)$, for every $\varepsilon >0$ there is a $N \geq 1$ such that, for every $k, l \geq N$, $\rho (x_k , x_l) < \varepsilon$, that, when applied to a space endowed with the discrete metric $d$, gives rise to the fact that there is a $N \geq 1$ such that, for every $k, l \geq N$, $d (x_k , x_l)=0$. Hence, $x_n$ converge, and $\mathbb{N}$ is complete.
As I wrote, the result is fairly obvious, but I was wondering about my way of writing it, and of course also about the content, i.e. the fact that I don't actually show how the limit looks like, but simply point out that it has to exist.
As always, any feedback regarding anything will be more than welcome.
Thank you for your time.
You should take out the second sentence. Instead, apply the definition of a Cauchy sequence to $(x_n)$, taking $\epsilon = 1/2$ in (or any other positive number less than $1$). The use the definition of the discrete metric to get $x_k = x_N$ for all $k \ge N$. This will show that your Cauchy sequence $(x_n)$ is eventually constant with limit $x_N$.