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 $|x_{n_k}-x|<{\varepsilon\over 2}$. Let $N_2$ be such that $n,m\ge N_2$ implies $|x_n-x_m|<{\varepsilon \over 2}$.

Then $n>N=\max(N_1,N_2)$ implies $$ |x_n-x|\le |x_n-x_N|+|x_N-x|<\varepsilon$$

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.

Here's a not-so-fancy way:

Let $\{a_n\}$ be a Cauchy sequence.

If the set of values in the (image of the) sequence is finite, then use the Cauchy criterion to show that the sequence is eventually constant (and so converges).

If the set of values of the sequence is infinite, then use compactness to finite a limit point of this set. Use this limit point to construct a convergent subsequence of the original sequence. Then use the Cauchy criterion to show the original sequence converges to the same limit as the subsequence.