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Let $(x_n)$ be a Cauchy sequence in a compact metric space $X$. By sequential compactness of a compact metric space, it has a convergent subsequence $x_{n_j}\to x\in X$. Given $\epsilon >0$, choose $N$ such that $d(x_m,x_n)<\epsilon /2$ whenever $m,n\ge N$, and then choose $j$ such that $n_j\ge N$ and $d(x_{n_j},x)<\epsilon /2$. For every $n\ge N$, the triangle inequality givesThus the whole sequence converges to $x$. Every Cauchy sequence converges in $X$, so $X$ is a complete metric space.

This is always true: it is the Heine-Cantor theorem. If uniform continuity failed, there would be an $\epsilon >0$ and sequences $x_n,y_n\in X$ such thatBy sequential compactness of a compact metric space, some subsequence $x_{n_k}$ converges to $x\in X$. The triangle inequality gives $y_{n_k}\to x$. Continuity then makes both image subsequences converge to $f(x)$, contradicting their separation by $\epsilon$.