Absolute Value Proof: if $-a \leq x \leq a$, then $|x| \leq a$. I want to prove the following proposition:

If $-a \leq x \leq a$, then $|x| \leq a$, where $x,a \in \mathbb{R}$.

Here's my proof:

By trichotomy, there are two possibilities: either $x \geq 0$, or $x<0$. If $x \geq 0$, then $|x| = x$, so $|x| \leq a$. If $x<0$, then $|x| = -x$. We note that $-x > a$ iff $x < -a$, but this is impossible because $x \geq -a$. It follows that $-x \leq a$, i.e. $|x| \leq a$.   

I think it works. I'd be pleased if someone could verify my work - or suggest a more effective way of proving the proposition. Thanks!
 A: I think you've got a typo (I hope it's only a simple typo; otherwise, you have a significant error in omitting a case!): Look carefully at your post: After considering $x\geq 0$, you then consider the case $x\gt 0$ for which you add that it follows that $|x| = -x$. That is only true if you meant to be considering the case $x\lt 0$.
If it is a simple typo, then your argument is fine.
ADDED AFTER EDIT: NOW you've got a tight argument!
A: Looks pretty good. Maybe some of the phrasing could be cleaned up; sort of weird to start by saying "By tri-chotomy, there are two possibilities..." but this is nit-picky.
I might have said:
"If $x \geq 0$, then $|x| = x \leq a$, where the first equality follows because $x$ is non-negative, and the subsequent inequality is given."
But most changes would be about as cosmetic as this.
(If you want more proof practice, you could try proving the contrapositive...)
A: It looks good!
As one alternative, for the $x<0$ part, we have $x=-|x|,$ so $-a\le-|x|,$ and so $|x|\le a.$
As another alternative, note that $x-a\le 0$ and $0\le x+a$,so $x^2-a^2\le0(x-a)=0,$ so $x^2\le a^2.$ Taking the square root on both sides preserves the inequality, since $t\mapsto\sqrt t$ is an increasing function on $[0,\infty),$ and squares of real numbers are non-negative, so we have $$\sqrt{x^2}\le\sqrt{a^2}.$$ Note that by assumption, $-a\le a,$ so $0\le 2a,$ so $0\le a,$ and so $$\sqrt{x^2}\le a.$$ Finally, if you know (or can prove) that $|x|=\sqrt{x^2}$ for all $x\in\Bbb R,$ then we have $$|x|\le a,$$ as desired.
