We seek to show that $f:K \to X$ is continuous with $K$ a compact metric space, then $f$ is uniformly continuous. Recall that for $f:K\to X$ , uniform continuity is the condition that for any $\varepsilon>0$ , there exists $\delta$ such that$$ d_K(x,y) < \delta \implies d_X (f(x),f(y)) < \epsilon$$ for all $x,y\in K$

Suppose $K$ is a compact metric space, $f$ continuous on $K$ . Let $\epsilon > 0$ . For each $k \in K$ choose $\delta_k$ such that $d(k,x) \leq \delta_k$ implies $d(f(k),f(x)) \leq \frac{\epsilon}{2}$ . Note that the collection of balls $B(k, \frac{\delta_k}{2} )$ covers $K$ , so by compactness there is a finite subcover, say involving $k_1, \ldots, k_n$ . Take \begin{equation*} \delta = \min_{i=1,\ldots,n} \frac{\delta_{k_i}}{2} \end{equation*}Then, suppose $d(x,y) \leq \delta$ . By the choice of $k_1,\ldots,k_n$ and the triangle inequality, there exists an $i$ such that $d(x,k_i),d(y,k_i) \leq \delta_{k_i}$ . Hence, \begin{eqnarray} d(f(x),f(y)) &\leq& d(f(x),f(k_i)) + d(f(y),f(k_i)) \\ &\leq& \frac{\epsilon}{2} + \frac{\epsilon}{2} \end{eqnarray} As $x,y$ were arbitrary, we have that $f$ is uniformly continuous.
This proof is similar to one found in Mathematical Principles of Analysis, Rudin.