Direct proof. Square root function uniformly continuous on $[0, \infty)$ (S.A. pp 119 4.4.8) (http://math.stanford.edu/~ksound/Math171S10/Hw8Sol_171.pdf)
Prove for all $e > 0,$ there exists $d > 0$ : for all $x, y \ge 0$, $|x - y| < d \implies |\sqrt{x} - \sqrt{y}| < e$. 
(a) Given $\epsilon>0$, pick $\delta=\epsilon^{2}$. First note that $|\sqrt{x}-\sqrt{y}|\leq|\sqrt{x}+\sqrt{y}|$.
Hence if $|x-y|<\delta=\epsilon^{2}$, then
$
|\sqrt{x}-\sqrt{y}|^{2}\leq|\sqrt{x}-\sqrt{y}||\sqrt{x}+\sqrt{y}|=|x-y|<\epsilon^{2}.
$
Hence $|\sqrt{x}-\sqrt{y}|<\epsilon$.
1. Where does $|\sqrt{x}-\sqrt{y}|\leq|\sqrt{x}+\sqrt{y}|$ issue from?  How to presage this presciently?
Yes...can prove it. Square both sides. By dint of $|a|^2 = (a)^2$:
$|\sqrt{x}-\sqrt{y}|^2 \leq|\sqrt{x}+\sqrt{y}|^2 \iff (\sqrt{x}-\sqrt{y})^2 \leq(\sqrt{x}+\sqrt{y})^2 \iff - 2\sqrt{xy} \le 2\sqrt{xy} \\ \iff 0 \le 4\sqrt{xy}. █ $.   
2. Figue or Intuition please for $|\sqrt{x}-\sqrt{y}|\leq|\sqrt{x}+\sqrt{y}|$? Feels fey.
I know $|x + y| \le |x| + |y| \iff |x - y| \le |x| + |y|$.
3. For scratch work, I started with $|\sqrt{x} - \sqrt{y}| < e$. But answer shows you need to start with $|\sqrt{x} - \sqrt{y}|^2$. How to presage this vaticly? What's the scratch work for finding $d = e^2$?
4. How does $d = e^2$ prove uniform continuity? I know this proves uniform  continuity.
Solution needs to prove $d = e^2$ doesn't depend on x or y? 

2. You know that $\sqrt{x}$ and $\sqrt{y}$ are non negative. The sum of two non-negative numbers is always at least as great as their difference. Alternatively,  $|x| + |y| = |x + y|$ for non negative $x$ and $y$. Thus $|\sqrt{x} - \sqrt{y}| \le |\sqrt{x}| + |\sqrt{y}| = |\sqrt{x} + \sqrt{y}|$.
3. Practise! The solution is not obvious.
4. $\delta$ doesn't depend on $x$ or $y$. It is a function solely of $\epsilon$. Thus, the same $\delta$ works for any $x$ and $y$, which is what is required for uniform continuity.
