Uniform continuity implies we just need a small enough $\delta$? Definition of uniform continuity of a function $f:D\to\mathbb{R}$: $$\forall\epsilon>0,\exists\delta>0:x,y\in D \wedge |x-y|<\delta\implies|f(x)-f(y)|<\epsilon.$$
Consider the function $f(x)=\sqrt x$ on the interval $[0,1]$. This is uniformly continuous, since our domain is compact. But for a function to be uniformly continuous, don't we need a small enough $\delta$ so that it can be used on our entire domain? If we were to approach $0$ from the left, wouldn't we need our $\delta$ smaller and smaller to no end? It seems that we would not ever find a small enough $\delta$ that can be used on our domain.
I do not care for a proof that $f$ is uniformly continuous; rather, I'm curious as to how we are able to find a small enough delta (unless, of course, I am misinterpreting the definition). Thanks!
 A: It's a good question. For $x \mapsto \sqrt{x}$, you'd typically choose something like $\delta = \min(1, \frac{1}{2} \epsilon^2)$. (I could be off by a small constant). The point is that this choice of $\delta$ is small enough to work near $0$ (which, as you correctly observe, is the tricky area of this function), but also suffices everywhere else. 
You may be used to the idea that "the way to choose $\delta$ is to divide $\epsilon$ by the derivative of $f$ at your point, perhaps with an extra factor of $2$ to be on the safe side," which is a decent first-cut strategy. But here, we've got a more clever way to pick $\delta$ -- in particular, it's not linear in $\epsilon$ -- and the wonderful thing is that it works both at the origin and elsewhere. 
The "divide delta by the derivative" approach works pretty well for many functions. For instance, if your function happens to satisfy $0 < |f'(x)| < M$ for every $x$, then something like $\delta = \epsilon / M$ will probably work to show uniform continuity. But as the example above shows, other techniques can be used even in cases where there's no similar bound on the derivative. 
