Is this proof for if $0 < a < b$ then $a^2 < b^2$ correct? I'm reading the book 'How to prove it' from Daniel Velleman which he presents a proof for the following; if $0 < a < b$ then  $a^2 < b^2$ as;
Proof. Suppose $0 < a < b$. Multiplying the inequality $a < b$ by the positive number $a$ we can conclude that $a^2 < ab$, and similarly multiplying by $b$ we get $ab < b^2$. Therfore $a^2 < ab < b^2$, as required. Thus if $0 < a < b$ then  $a^2 < b^2$.
However, I was also wondering if the statement could be proved using the following method.
Proof. Suppose that $0 < a < b$. Taking the square root of both sides of the inequality $\sqrt{a^2} < \sqrt{b^2}$ we get our original hypothesis $a < b$ . Thus if $0 < a < b$ then  $a^2 < b^2$. 

In the proof from the book you presented, you've assumed $0<a<b$ and deduced $a^2<b^2$.
In your presented proof, you've essentially assumed $a^2<b^2$ and deduced $a<b$. Why? The hypothesis $0<a<b$ is never used and by taking the square root of $a^2<b^2$, you implicitly assume that statement, deducing $a<b$ from it.
Thus, conceptually you have shown $B\Rightarrow A$ for the corresponding statements $A,B$, i.e. you have established $A\Leftrightarrow B$ using $A\Rightarrow B$ from the previous proof.
EDIT: As discussed with Ennar in the comments, there are additionally some issues with deducing $a<b$ from $a^2<b^2$ as it requires the square root function to be monotone, a property that follows from the fact the the square function is monotone on $[0,\infty)$. Then of course, you can not establish $B\Rightarrow A$ without first establishing $A\Rightarrow B$, i.e. there is another circularity of reasoning.
As you can see, there are even more implicit assumptions than I had initially pointed out.
