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

Suppose $f$ is not uniformly continuous. Then

In particular by letting $\delta=1/k$ we can construct two sequences $x_{k}$ and $y_{k}$ such that

Since $X$ is compact the two sequence have convergent subsequences i.e.

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 $d(f(x_{k}),f(y_{k}))\geq\epsilon$.