If $x$ is a real number, then $|x+1| \leq 3$ implies that $-4 \leq x \leq 2$.
I've tried to prove this by exhaustion, is that the right way to prove it?
You can use the definition of $|.|$.
The absolute value of real number $a$ (we'll write it $|a|$) it's the distance between the point,
which corresponds to $a$ in the $x$-axis and the origin.
In our case the distance between $x+1$ and $0$ less or equal to $3$.
Thus, we have $$-3\leq x+1\leq3$$ or $$-4\leq x\leq2$$
I'd say that the easiest way to prove this is to simply solve the inequality for x.
We know that:
|x+1| = 3, simplifies to:
x+1 = +3 AND x+1=-3
So by this logic, we can say that:
x+1 ≤ 3 AND x+1 ≥ -3 (switching the ≤ to ≥ when we change the sign on the 3)
x ≤ 2 AND x ≥ -4 (subtracting 1 in on both sides)
Then we can just simplify this expression down into the one, written as, -4 ≤ x ≤ 2, just as you needed to show.
Hope this helps. This is just the easiest way to show it, though there are many others.