Inequality $ \vert \sqrt{a}-\sqrt{b} \vert \leq \sqrt{ \vert a -b \vert } $

This is not going to be a polished argument. Rather, I’m presenting it as I found it, as an illustration of how you might attack such a problem.

Since the absolute value and the square root are non-negative, $|\sqrt{a}-\sqrt{b}|\le\sqrt{|a-b|}$ if and only if $\left(\sqrt{a}-\sqrt{b}\right)^2\le|a-b|$, i.e., if and only if $a-2\sqrt{ab}+b\le|a-b|$. Without loss of generality we may assume that $a\ge b$ and ask whether $a+b-2\sqrt{ab}\le a-b$. Evidently this is the case if and only if $2b\le2\sqrt{ab}$, or $b\le\sqrt{ab}$. This is certainly true if $b\le 0$. Since we’re assuming that $a\ge b$, it’s also true if $b>0$.

$$|a-b|=|(\sqrt a-\sqrt b)(\sqrt a+\sqrt b)|=|\sqrt a-\sqrt b|\cdot|\sqrt a+\sqrt b|\ge |\sqrt a-\sqrt b|\cdot|\sqrt a-\sqrt b|$$