Proof
(related to Theorem: Continuous Real Functions on Closed Intervals are Uniformly Continuous)
 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 $xy < \delta.$
 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 $$x_ny_n < \frac 1n\quad\text{ and }\quad f(x_n)f(y_n)\ge \epsilon.$$
 $(x_n)_{n > 0}$ 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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References
Bibliography
 Forster Otto: "Analysis 1, Differential und Integralrechnung einer VerĂ¤nderlichen", Vieweg Studium, 1983