Let $X$ be a compact metric space. Prove that $X$ is also complete.
Here's a more direct way:
Let $(x_n)_{n \in \mathbb{N}}$ a cauchy sequence in $X$. That is, for any $\epsilon /2 >0$ there exists $N_0 \in \mathbb{N}$ such that $d(x_n,x_m)<\epsilon /2$ for $n,m \geq N_0$.
$X$ is compact, that is that $(x_n)_{n \in \mathbb{N}}$ has a convergent sub-sequence, that is $\lim_{n\rightarrow ∞}(x_{n_k})_{k \in \mathbb{N}}=x_0$, ir that for any $\epsilon /2 >0$ there exists $N_1 \in \mathbb{N}$ such that $d(x_{n_k},x_0)<\epsilon /2$ for $k \geq N_1$. Note that $k>n_k$ should be satisfied.
So we want the cauchy sequence to be convergent, that is for any $\epsilon >0$ there exists $N \in \mathbb{N}$ such that $d(x_n,x_0)<\epsilon $ for $n \geq N$.
Taking $N=max(N_0,N_1)$ and by triangle inequality we have: $d(x_n,x_0) \leq d(x_n,x_{n_k}) + d(x_{n_k},x_0) \rightarrow d(x_n,x_0) < \epsilon /2 + \epsilon /2 = \epsilon$
Is it correct?
Your proof is o.k.
But you should write $N=max (N_0,N_1) $