Prove $||a|-|b| | \leq |a-b|$

Two small comments:

In case three: consider $a \ge 0; b< 0$ (otherwise you haven't covered the case where $a =0; b < 0$.)

An also in case three: You state $|a| + |b| \le a - b$. It's stronger to state $|a| + |b| = a-b$.

Other than that this is fine.

However in case 3) you refer to an earlier result $|a+b| \le |a| + |b|$. You should either explicitly mention that result or try to prove without it.

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As an alternative prove as you have already proven $|a + b| \le |a|+|b|$ you might consider this slick proof:

$|a| = |a+b -b| \le |a-b| + |b|$ so $|a| - |b| \le |a-b|$.

Likewise $|b| = |a+b - a| \le |b-a| + |a|$ so $|b| - |a| \le |b-a| = |a-b|$

So $||a| -|b|| = \pm(|a| - |b|) \le |a-b|$.

Observe $a.b \leq |a||b|$ then $|a-b|^2=(a-b)^2= a^2-2a.b+b^2 \geq |a|^2 -2|a||b| +|b|^2=(|a|-|b|)^2$ $\Rightarrow |a-b| \geq ||a|-|b||$

Suppose $a,b \in X,$ where $X$ is a normed linear space. By the triangle inequality, $|a| \le |a-b|+|b|,$ which implies $|a|-|b| \le |a-b|.$ Similarly $|b|-|a| \le |b-a|=|a-b|.$ The conclusion follows.