A question of approach.. really basic I am afraid.
Given $$|a|\leq M$$ is this the correct approach?
$$\left| a \right| \leq M = \begin{cases} a \leq M, & a \geq 0 \\ -a \leq M, & a \leq 0 \end{cases}$$
$$ \qquad = \begin{cases} a \leq M, & a \geq 0 \\ a \geq -M, & a \leq 0 \end{cases} $$
thus $$-M \leq a \leq M$$
Thanks.
The logic/reasoning is correct; the notation can be improved.
The use of the "$=$" sign is a bit problematic; these are equivalent statements, but we usually use another symbol, $\iff$ (if and only if, or "iff"), to denote equivalent statements.
As for the intervals for $a$, it would be cleaner if they were separate from eachother, as they currently both have $0$ ($a\leq0$ and $a\geq0$). Instead, you should include zero in only one of them.
Applying all this would result in your approach being written as: $$\left| a \right| \leq M \iff \begin{cases} a \leq M, & a \geq 0 \\ -a \leq M, & a< 0 \end{cases}\iff \begin{cases} a \leq M, & a \geq 0 \\ a \geq -M, & a < 0 \end{cases}$$ thus $-M\leq a \leq M$.