Prove that $| |a|-|b| | \le |a-b|$
(1) I square both sides since I know they are positive:
$$a^2-2|ab|+b^2 \le a^2-2ab+b^2$$
$$|ab| \ge ab$$
Let $m=ab$
$$|m| \ge m$$
Which is part of the definition of the absolute value.
Is it enough or I need to rewrite my proof?
It follows from the inequalities $$|a|=|a-b+b|\le |a-b|+|b|\implies |a|-|b|\le |a-b|.$$ Because of symmetry it is
$$|b|-|a|\le |a-b|.$$ Thus you are done.