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:

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