Prove that if $X$ is compact, then every continuous function $f: X \to Y$ is uniformly continuous.
What I did was proof by contradiction, so I assumed $f$ was not uniformly continuous. Then, there exists $\epsilon > 0$ and $\{x_n\}$ and $\{y_n\}$ in $X$ such that $d\bigl(f(x_n), f(y_n)\bigr) \geq \epsilon$ for all $n$, but $$ \lim_{n \to \infty} d(x_n, y_n) = 0. $$
Since $X$ is compact, there is a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ that converges to some $x \in X$. Since the limit is zero, $\{y_{n_k}\}$ also converges to $x$. Since $f$ is continuous at $x$, $$ \lim_{k \to \infty} f(x_{n_k}) = f(x) \quad\text{and}\quad \lim_{k \to \infty} f(y_{n_k}) = f(x) $$ which gives: $$ d\bigl(f(x_{n_k}), f(y_{n_k})\bigr) \leq d\bigl(f(x_{n_k}), f(x)\bigr) + d\bigl(f(x), f(y_{n_k})\bigr) $$ which converges to $0$ as $k \to \infty$ since $f$ is continuous, which contradicts the fact that $d\bigl(f(x_n), f(y_n)\bigr) \geq \epsilon$ for all $n$. Thus, $f$ must be uniformly continuous.
Is this logic right? Is it missing something?
Your proof looks good.
Here I will provide you with a direct proof, which may be more common in textbooks.
Given $\varepsilon>0.$ By continuity, $\exists B(p,\delta_p)\subseteq X$ such that $$d(f(x),f(p))<\varepsilon \quad \text{ if }\quad x\in B(p,\delta_p)$$ where the notation $B(p,\delta_p)$ here means the open ball centered at $p$ with radius $\delta_p>0.$
Then we see $X\subseteq\bigcup_{p\in X} B(p,\frac12\delta_p)$. Since $X$ is compact we can write $$X=\bigcup_{j=1}^m B(p_j,\frac12\delta_{p_j})$$
Choose $\delta=\min_{j=1,\cdots,m}\{\delta_{p_j}\}>0.$
(edit: Should be $\delta=\min_{j=1,\cdots,m}\{\frac12\delta_{p_j}\}>0,$ see the comment below.)
Thus for any $x,y\in X$ such that $d(x,y)<\delta$, we have $$x\in B(p_j,\frac12\delta_{p_j})\text{ for some }j$$ and $$y\in B(x,\frac12\delta_{p_j})$$ meaning that both $x,y\in B(p_j,\frac12\delta_{p_j})$.
Hence $$d(f(x),f(y))\leq d(f(x),f(p_j))+d(f(p_j),f(y))<2\varepsilon$$