Proof

Let $[a,b]$ be a closed real interval, $\mathbb R$ be the set of real numbers and $f:[a,b]\mapsto \mathbb R$ a continuous function.
We will prove by contradiction to that $f$ is uniformly continuous, i.e. for every $\epsilon > 0$ there is a $\delta > 0$ such that $|f(x)-f(y)| < \epsilon$ for all $x,y\in D$ with $|x-y| < \delta.$
- Assume, $f$ is not uniformly continuous.
- Then there is an $\epsilon > 0$ such that for every natural number $n > 0$ such that there as some points $x_n, y_n\in [a,b]$ with $$|x_n-y_n| < \frac 1n\quad\text{ and }\quad |f(x_n)-f(y_n)|\ge \epsilon.$$
- $(x_n)_{n > 0}$ is bounded real sequence.
- Because every bounded real sequence has a convergent subsequence, so has $(x_n)_{n > 0}.$
- Let this subsequence be $(x_{n_k})_{k\in\mathbb N}$, and let its limit be $\lim_{k\to\infty} x_{n_k}=p.$
- Since convergence preserves upper and lower bounds, and since $a \le x_{n_k} \le b$ for all $k\in\mathbb N$, we have that $p\in[a,b].$
- Since $|x_{n_k}-y_{n_k}| < \frac 1{n_k}$ by construction, we have also that $\lim_{k\to\infty} y_{n_k}=p.$
- From the definition of continuity of $f$ it follows $$\lim_{k\to\infty}\left(f(x_{n_k})-f(y_{n_k})\right)=f(p)-f(p)=0.$$
- This is a contradiction to $|f(x_{n_k})-f(y_{n_k})|\ge \epsilon$ for all $k\in\mathbb N.$
- Thus the assumption, $f$ was not uniformly continuous must be wrong.
- Therefore $f$ is uniformly continuous.
  ∎

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Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983