Proof of Heine-Cantor: Continuity on Compact Interval Implies Uniform Continuity

Let $f:[a,b]\to\mathbb R$ be continuous. If $f$ were not uniformly continuous, there would exist $\varepsilon_0>0$ and sequences $x_n,y_n\in[a,b]$ with $|x_n-y_n|\to0$ but $|f(x_n)-f(y_n)|\ge\varepsilon_0$. By compactness of $[a,b]$, after passing to a subsequence $x_n\to x_\ast\in[a,b]$. Then $y_n\to x_\ast$ as well since $|x_n-y_n|\to0$. Continuity of $f$ gives $f(x_n)\to f(x_\ast)$ and $f(y_n)\to f(x_\ast)$, hence $|f(x_n)-f(y_n)|\to0$, contradicting $|f(x_n)-f(y_n)|\ge\varepsilon_0$. So $f$ is uniformly continuous.