Every compact metric space is complete

If you do not wish to use the Heine-Borel theorem for metric spaces (as suggested in the answer by Igor Rivin) then here is another way of proving that a compact metric space is complete:

Note that in metric spaces the notions of compactness and sequential compactness coincide. Let $x_n$ be a Cauchy sequence in the metric space $X$. Since $X$ is sequentially compact there is a convergent subsequence $x_{n_k}\to x \in X$.

All that now remains to be shown is that $x_n \to x$. Since $x_{n_k}\to x$ there is $N_1$ with $n_k \ge N_1$ implies $d(x_{n_k},x)<{\varepsilon\over 2}$. Let $N_2$ be such that $n,m\ge N_2$ implies $d(x_n,x_m)<{\varepsilon \over 2}$.

Let $n\geq N=\max(N_1,N_2)$ and pick some $n_k\geq N$. Then $$ d(x_n,x)\le d(x_n,x_{n_k})+d(x_{n_k},x)<\varepsilon.$$

Hence $X$ is complete.

Let $\langle F_n\rangle_{n\in\Bbb{N}}$ be a descending sequence of nonempty closed sets satisfy that $\operatorname{diam} F_n\to 0$ as $n\to\infty$. You can easily check that if $m_1<m_2<\cdots<m_k$ then $$ F_{m_1}\cap F_{m_2}\cap\cdots\cap F_{m_k} =F_{m_k}\neq \varnothing $$ so $\langle F_n\rangle_{n\in\Bbb{N}}$ satisfies finite intersection property. Since $(X,d)$ is a compact metric space, $\bigcap_{n\in\Bbb{N}} F_n$ is not empty. Since $$ \operatorname{diam} \bigcap_{n\in\Bbb{N}} F_n \le \operatorname{diam} F_m\to 0\qquad \text{as }\> m\to\infty $$ so $\bigcap_{n\in\Bbb{N}} F_n$ contains at most one point. So $\bigcap_{n\in\Bbb{N}} F_n$ is singleton.