Compactly supported continuous function is uniformly continuous

Let us first see what is continuity. Given a positive $\epsilon$, continuity of $f$ means that at each $x$, we can find a small $\delta$ such that $x'\in (x-\delta,x+\delta)$ implies $f(x')\in (f(x)-\epsilon,f(x)+\epsilon)$. So around each $x$, we have a small open set that is mapped very close to $f(x)$. Also note that points in the same small open set are mapped to points close to each other, since they are both mapped to points close to $f(x)$. But the problem here is that this radius of the open sets $\delta$ depends on the point $x$.

Now compactness means you can find a finite collection of such small open sets that they cover the support of the function, and this is good enough for us. Because among these finitely many open sets, you can determine the smallest radius, say $r$. If the distance between $x$ and $x'$ are less than $1/10r$, then you know they must lie in the same open set, then $f(x)$ and $f(x')$ must be close. Now this $r$ does not depend on $x$, you have uniform continuity.