I'm not familiar with proofs and I dunno if I am doing them right. Here's one from Spivak's Calculus: 14. (a) Prove that |a| = |-a| (The trick is not to become confused by too many cases.) So I did: $$ |x| = \left\{\begin{aligned} &x &&: x \ge 0\\ &-x &&: x < 0 \end{aligned} \right.$$
Case 1: $a \ge 0$
Then $|a| = a$ by definition.
Impliying |a| = |a|
Case 2: $a < 0$
Then $|a| = -a$ by definition.
Implying $|a| = |-a|$
Thus $|a| = |-a|$ if $x \in \mathbb{R}$.
It's not quite correct. You've concluded that $|a| = |a|$, which is correct but not what you wanted in case 1. Your last line is also something about $x$ being a real number, but you intended $a$ there.
Generally, though, you'll want to write it in a bit more reader-friendly method, explaining what you're doing and defining your objects; something like
Let $a$ be a real number; we break the proof up into two cases: $a \ge 0$ and $a < 0$. If $a \ge 0$, then by definition of absolute value, $|a| = a$. On the other hand, $-a \le 0$, so $|-a| = -(-a) = a$, giving the desired result. Next if $a < 0$, ....