I wish to show that $\sqrt{x}$ is uniformly continuous for all $x \geq 0$.
Fix $\epsilon > 0$. Let $\delta = \epsilon$. Then $|x - y| < \delta$.
Then $|\sqrt x - \sqrt y| < |\sqrt x - \sqrt y| |\sqrt x + \sqrt y| = |x - y| < \delta = \epsilon$, so $|\sqrt x - \sqrt y| < \epsilon$.
This seems a little too simple, so I'm not sure if I've done it correctly. Any feedback?
Hint:
$$|\sqrt{x} - \sqrt{y}|^2 \leqslant |\sqrt{x} - \sqrt{y}||\sqrt{x} + \sqrt{y}| = |x-y| \\ \implies |\sqrt{x} - \sqrt{y}| \leqslant \sqrt{ \ldots} $$