We prove this theorem in the case when $X$ and $Y$ are metric spaces.

Suppose $f$ is not uniformly continuous. Then $$ \exists \epsilon>0\ \forall \delta>0\ \exists x,y\in X \quad d(x,y)< \delta \ \mathrm{but}\ d(f(x),f(y))\ge \epsilon. $$ In particular by letting $\delta=1/k$ we can construct two sequences $x_k$ and $y_k$ such that $$ d(x_k,y_k) < 1/k\ \mathrm{and}\ d(f(x_k),f(y_k)\ge \epsilon. $$

Since $X$ is compact the two sequence have convergent subsequences i.e. $$ x_{k_j} \to \bar x \in X, \quad y_{k_j} \to \bar y \in X. $$ Since $d(x_k,y_k)\to 0$ we have $\bar x = \bar y$ . Being $f$ continuous we hence conclude $d(f(x_{k_j}),f(y_{k_j})) \to 0$ which is a contradiction being $d(f(x_k),f(y_k))\ge \epsilon$ .