I've modified the proof. I request you to verify this proof.
CLAIM: If $X$ is a compact metric space and $f$ is continuous on $X$, then $f$ is uniformly continuous on $X$.
PROOF: If $f$ is not uniformly continuous on $X$, then there is a Cauchy sequence $(x_n)$ in X such that $(f(x_n))$ is not Cauchy. Since $X$ is compact(hence complete) $(x_n)$ converges to some point in $X$. Since $f$ is continuous, $(f(x_{n}))$ converges. But this would imply that $(f(x_n))$ is Cauchy which is a contradiction.
If $f$ is not uniformly convergent, there is some $\varepsilon >0$ such that for all $\delta =\frac1n>0$ there are $x_n, x'_n \in X$ such that $d(x_n, x'_n) < \frac1n$ and $d(f(x(\delta)), f(x'(\delta)) \ge \varepsilon$.
Then by (sequential) compactness of $X$, there are $x_0, x'_0 \in X$ and a series of indices $n_k$ such that $x_{n_k} \to x_0, x'_{n_k} \to x'_0 $ as $k \to \infty$.
As $d(x_{n_k}, x'_{n_k}) < \frac{1}{n_k}$ we see that in fact $x_0 = x'_0$ and then we have a contradiction with continuity of $f$ at $x_0$, as $d(f(x_{n_k}), f(x'_{n_k})) \ge \varepsilon$.