Can anyone tell me if this proof for the uniform continuity of $\sqrt{x}$ works?
Let $\varepsilon > 0$, $\delta = \varepsilon^2$, and let $x,u \in \left[0,\infty\right[$ with $|x - u| < \delta$. WLOG let $x \geq u$. Then, $|\sqrt x - \sqrt u| = \sqrt x - \sqrt u < \sqrt{u + \delta} - \sqrt{u} = \frac{\delta}{\sqrt{u + \delta} + \sqrt u} < \frac{\delta}{\sqrt \delta} = \sqrt\delta = \varepsilon$.
Many of the proofs I've seen involve checking uniform continuity on an interval like $[0,2]$ and on $[1, \infty[$, but this takes care of all cases at once, or is there something wrong?
Thanks a lot.
Yes, this works. Alternatively, one way to bypass the "WLOG" part and the whole fraction business is to leave everything squared:
$$|\sqrt x - \sqrt u|^2 \leq |\sqrt x - \sqrt u||\sqrt x + \sqrt u| = |x-u| < \epsilon^2 \implies |\sqrt x - \sqrt u| < \epsilon. $$
I think this is pretty much the most streamlined proof you can find of this fact, and you are quite right in saying that you don't have to go through the trouble of invoking continuity on a compact interval and then Lipschitz continuity on $[1,\infty)$ to prove uniform continuity of $\sqrt x$. (Of course, these techniques are important and applicable to a much larger class of functions, and this is a nice first example to typically apply them to.)