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? 
 A: When $0\leq x\leq y$ then
$$\left(\sqrt{\mathstrut y}-\sqrt{\mathstrut x}\right)^2=y-2\sqrt{\mathstrut xy}+x\leq y-x\ .$$
It follows that
$$\sqrt{\mathstrut y}-\sqrt{\mathstrut x}\leq\sqrt{\mathstrut y- x}\ .$$
This implies
$$\left|\sqrt{\mathstrut y}-\sqrt{\mathstrut x}\right|\ \leq\ \sqrt{|y-x|}$$
for arbitrary $x$, $y\in{\mathbb R}_{\geq0}$ and therewith the uniform continuity of $\sqrt{\cdot}\>$. 
A: Or just prove using the binomial theorem that for $0\le x\le y=x+h$, you get
$$\sqrt[n]{x+h}\le\sqrt[n]{x}+\sqrt[n]h,$$
so that finally for any $x,y\ge 0$
$$\left|\sqrt[n]y-\sqrt[n]x\right|\le\sqrt[n]{|y-x|}.$$

Now compare with the definition of Hölder continuity with Hölder index $α\in(0,1]$,
$$|f(y)-f(x)|\le C\cdot|y-x|^\alpha$$
and use that Hölder continuity implies uniform continuity.

@1) already answered sufficiently in question.
@2) The use of the binomial theorem for an arbitrary power $n$, essentially $A^n+B^n\le(A+B)^n$, with $A=\sqrt[n]x$ and $B=\sqrt[n]{y-x}$ generalizes the method given in the question.
@3) In this generalized form as equation for Hölder continuity, the motivation is much clearer.
@4) choosing $δ=ϵ^n$ results in the conclusion that if $|x-y|<δ$ for any $x,y\ge 0$, then $\left|\sqrt[n]y-\sqrt[n]x\right|<\sqrt[n]{ϵ^n}=ϵ$, and this proves uniform continuity.
A: 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.
