$\newcommand{\sgn}{\operatorname{sgn}}$It was left as an exercise to show that:
$\forall a,b\in\Bbb R,$ $$\left|\sgn(a)|a|^{1/2}-\sgn(b)|b|^{1/2}\right|^2\le2\cdot|a-b|$$
The only techniques I have seen used to solve these problems (they seem to occur often in the functional analysis of $\mathcal{L}^p$ spaces - I've seen this method used in the proofs of Banach-Saks and some other theorems) is to do a division and show that the resulting rational function is bounded and positive in the region of interest by taking limits. My solution given below works, but it felt like more work than should have been necessary for this problem - I am asking if there are simpler approaches.
I then am reducing this problem to showing that:
$$\frac{|\sgn(a)|a|^{1/2}-\sgn(b)|b|^{1/2}|^2 }{|a-b|}\le2$$
Always. Since the expression is symmetric in $a,b$, assume wlog that $a\ge b$. Note of course that the expression is always positive, so we need only show that it is bounded from above.
Consider then the above as a function $f(a)$ defined on regions where $a\ge b$. If $b=0$ the expression is clearly $1\le 2$ for all $a$, so instead take $b\gt 0$ as a first case. For $a\ge b\gt 0$:
$$f(a)=\frac{a+b-2\sqrt{ab}}{a-b}=\frac{(a^{1/2}-b^{1/2})^2}{(a^{1/2}-b^{1/2})(a^{1/2}+b^{1/2})}=\frac{a^{1/2}-b^{1/2}}{a^{1/2}+b^{1/2}}$$
As $a\to b^+$ the limit is $0\le 2$; now take its derivative:
$$f'(a\ge b\gt0)=\frac{\sqrt{b/a}}{a+b+2\sqrt{ab}}\gt0$$
So as $f'$ is continuous we have that $f$ is a strictly increasing function, beginning (in the range $a\ge b\gt0$) at $0$ and has its upper bound at $\lim_{a\to\infty}f(a)=1\lt2$. Therefore $0\le f\lt 2$ when $a\ge b\ge0$.
Now suppose $b\lt0$. There are two cases; $a\gt0,b\le a\lt 0$. $a=0$ is excluded as the function is clearly $1$ in this case also. First take $a\gt0$:
$$f(a)=\frac{a-b+2\sqrt{a|b|}}{a-b}\ge1$$
And: $$f'(a)=\frac{(1+\sqrt{|b|/a})(a-b)-a+b-2\sqrt{a|b|}}{(a-b)^2}=\frac{-b\sqrt{|b|/a}-\sqrt{a|b|}}{(a-b)^2}$$
Let the numerator be $h(a)$. $h'(a)=-\frac{b}{2a}\sqrt{\frac{|b|}{a}}-\frac{1}{2}\sqrt{\frac{|b|}{a}}=0\iff a=-b$, and as $a$ exceeds this point $h'(a)\lt 0$ clearly, so this is a maximum of $h$, and $f(-b)=2$ and thus when $a\gt0,\,f\le 2$.
Now suppose $b\le a\lt 0$:
$$f(a)=\frac{a+b+2\sqrt{|a||b|}}{b-a}$$
And: $$f'(a)=\frac{(1-\sqrt{|b|/|a|})(b-a)+a+b+2\sqrt{|a||b|}}{(b-a)^2}=\frac{2b+\sqrt{|a||b|}-b\sqrt{|b|/|a|}}{(b-a)^2}$$
Call the numerator $g(a)$. $g'(a)=-\frac{1}{2}\sqrt{|b|/|a|}-\frac{b}{2|a|}\sqrt{|b|/|a|}\gt0$ since $|b|\gt|a|$. Therefore $g$ is increasing; take $\lim_{a\to b^+}g(a)=0$, which implies that $g$ increases beyond $0$ and is always positive. Thus $f'$ is positive and $f$ increases. As $a\to 0^-$, $f\to 1$, and this implies that $f\le 1$ in the region $b\le a\le0$.
Having considered all cases for $a\ge b$ wlog, this concludes that $f\le 2$.
This was tedious work for me; I made many mistakes with absolute value derivatives along the way... was there a more elementary way of seeing this inequality? Royden leaves it as an exercise and I'm not great with inequalities - I presume one did not need to take this calculus approach.
Note: it is left as another exercise to show that:
$$\left|\sgn(a)\cdot|a|^2-\sgn(b)\cdot|b|^2\right|\le2\cdot|a-b|(|a|+|b|)$$
Which definitely suggests these problems have a simpler solution, since calculating the above using my calculus method would be very tedious.
When in doubt, try to remove square roots.
Outline
If $u=|a|^{1/2}, v=|b|^{1/2},$ then the left side is either $(u+v)^2,$ if the signs of $a,b$ are different, or $(u-v)^2,$ if the signs are the two are the same.
You can treat the cases where $a$ or $b$ or both are zero as the same sign or different. Works either way.
Similarly, $|a-b|$ is $u^2+v^2$ if the signs are different, and $\left|u^2-v^2\right|$ if the signs are the same.
So you need to prove two inequalities for $u,v$ non-negative reals.
$$(u-v)^2\leq 2\left|u^2-v^2\right|\\ (u+v)^2\leq 2(u^2+v^2)$$
These are each not hard to prove.