Being that the proof that |x|≤ a ↔ -a ≤ x ≤ a became a bit tangled, I’ve written out the details.

First we show |x|≤ a → -a ≤ x ≤ a

Suppose |x|≤ a . Then since |x| ≥ 0, we have a ≥ 0. And –a ≤ 0. (This part of the proof does not depend on the sign of x.)

Now suppose x≥0, |x| = x . So we have -a ≤ 0 ≤ x = |x| ≤ a. Thus: -a ≤ x ≤ a.

OTOH if x<0, then x < a (since a ≥ 0) . And since |x| ≤  a, we have –x ≤ a, from which -a≤ x and again:

-a ≤ x ≤ a

Second we show -a ≤ x ≤ a → |x|≤ a

If x ≥ 0 then |x|=x and the 2^nd inequality in the hypothesis applies so that |x|≤ a . If x<0, then -x>0 and|x|= -x. But then -a ≤ x implies that -x≤a and thus again

|x|≤ a