How do I show a function $f:\mathbb{R}\to \mathbb{R}$ is uniformly continuous on a closed interval $[a,b]$?
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Use that the closed interval is compact:
Assuming continuity of $f$, for an $\varepsilon>0$ there are $\delta_x>0$ (depending on $x$) in each $x\in [a,b]$, such that with the open interval $I_x:=(x-\delta_x,\,x+\delta_x)$, we have $f(I_x)\ \subseteq \ (f(x)-\varepsilon,\ f(x)+\varepsilon)$.
Now, by compactness, we can choose finitely many of these intervals $I_x$ such that their union contains all points of $[a,b]$. Say, $I_{x_1},\dots,I_{x_k}$ are the chosen one, then you can define one $\delta$, independent from $x$ which will be good: $$\delta:=\min(\delta_{x_1},\dots,\delta_{x_k})\ /\,2$$ (Note that $\delta>0$ because there are only finitely many terms.)
Berci
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