Show that that $|\sqrt{x}-\sqrt{y}| \le \sqrt{|x-y|}$

Assume $x\ge y\ge 0$. Then we can do away with the absolute value signs. With that we get, from the original inequality, that $$ \sqrt x - \sqrt y \leq \sqrt{x-y} $$ Squaring both sides, we get $$ x -2\sqrt{xy} + y \leq x-y \quad \Rightarrow \quad 2\sqrt{xy}\geq 2y $$ which follows immediately from the assumption that $x\geq y$

HINT: look at the square the inequality, that is, prove:

$$|\sqrt x - \sqrt y|^2\leq|x-y|$$

We know $x,y\geq 0$. Hence, $x-y=(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})$ We exclude the trivial case $x=y$.

This implies

$$|\sqrt{x}-\sqrt{y}|\leq \sqrt{|\sqrt{x}-\sqrt{y}| |\sqrt{x}+\sqrt{y}|},$$

dividing by $\sqrt{|\sqrt{x}-\sqrt{y}|}$ will result in

$$\sqrt{|\sqrt{x}-\sqrt{y}|}\leq \sqrt{|\sqrt{x}+\sqrt{y}|}.$$

The last inequality is trivially true.