MHB Proving an absolute value inequality

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The discussion focuses on proving the inequality if |a| ≤ b, then -b ≤ a ≤ b, for real numbers a and b. The definition of absolute value is clarified, distinguishing cases where a is non-negative and negative. In Case I, it is noted that if a ≥ 0, then |a| = a, leading to a ≤ b, while in Case II, if a < 0, then |a| = -a, resulting in -b < 0 ≤ a. Participants express confusion regarding the logical structure of the proof, particularly the incorrect assertion that a > b in Case I. The need for a clearer, more coherent proof is emphasized.
cbarker1
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If $\left| a \right| \le b$, then $-b\le a\le b$.
Let $a,b \in\Bbb{R}$ The definition of the absolute value is $ \left| x \right|= x, x\ge 0$ and $\left| x \right|=-x, x< 0$, where x is some real number.

Case I:$a\ge 0$, $\left| a \right|=a>b$

Case II: a<0, $\left| a \right|=-a<b$the solution is $-b<0\le a\le b$

I work on a number line. yet I still have trouble with the proof.
 
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At which point of the proof are you facing difficulties?
 
Well, I, for one, have trouble understanding the proof since I don't follow its logical structure, and not just because $a>b$ should be $a\le b$ in case I. Maybe someone can write a more coherent proof.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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