If $0 \le a \lt b$ prove that $a^2 < b^2$

You don't need cases. Just say:

$a^2 - b^2 = (a - b)(a + b)$.

Since $a \ge 0$ and $ b>0$, $a + b > 0$ and $a - b < 0$ since $a < b$, thus: $(a -b)(a +b) < 0$. So $a^2 - b^2 < 0$. Thus $a^2 < b^2$.

If both are positive, then: $$a < b \implies \begin{cases} a^2 \leq ab \\ ab < b^2\end{cases} \implies a^2<b^2,$$ where we use transitivity in the last step.

Your proof of the case for $a=0$ is correct. Your case for the proof of $a>0$ is incorrect, because on the second line you assume what you are trying to show. Here is a way to do away with cases.$$\begin{align} 0 \le a \lt b \implies b-a>0 \\ \implies (b-a)^2>0 \\ \implies b^2-2ab+a^2>0 \\ \implies 2ab<a^2+b^2 \\ \implies 2a(a)<a^2+b^2 \quad \text{by assumption that} \quad a<b \\ \implies 2a^2<a^2+b^2 \\ \implies a^2<b^2 \end{align}$$

Hint: You could use a Calculus approach.

Let $f: [0,+\infty) \to \mathbb R $ defined by $f(x) = x^2$ and notice that $f'(x) = 2x \geq 0$. The function $f$ is non-decreasing.

If $a=0$, then it is trivial since then $b^2>0=a$

Assume $0<a<b$ then:

$aa<ba<bb$.

$$a^2-b^2 =\underbrace{(a-b)}_{<0}\underbrace{(a+b)}_{>0} <0 \implies a^2< b^2$$

$b^2 - a^2 \geq b^2 - ba$ since $a^2 \geq ba$. ("We're subtracting something bigger, so the total becomes smaller")

$b^2 - ba = b(b-a) > 0$ as each factor is positive.

Putting it all together: $b^2 - a^2 \geq b^2 - ba = b(b-a) > 0$.