(Theorem) Every compact metric space is complete

Theorem: Every compact metric space is complete.

Proof from Topology.

Let $(X,d)$ be a compact metric space and let $(x_n{)}_{n\in \mathbb{N}}$ be a Cauchy sequence. By sequential compactness, there exists a convergent subsequence $x_{n_i}\to x$. We want to show that the entire sequence has $x_n\to x$.

Given $ϵ>0$, convergence of the subsequence implies there exists $N\in \mathbb{N}$ such that if $i\ge N$, then $d(x_{n_i},x)<ϵ/2$, and Cauchy-ness of the overall sequence implies there exists $M\in \mathbb{N}$ such that if $n,m\ge M$, then $d(x_n,x_m)<ϵ/2$. Then if $k\ge K=\max (N,M)$, pick any $n_i\ge K$. Then

$$d(x_n,x)\le d(x_n,x_{n_i})+d(x_{n_i},x_n)=ϵ/2+ϵ/2=ϵ.$$