Solution verification: showing $|x|^2 = x^2$ and $|x|<|y| \iff x^2 < y^2$ I have two propositions to prove and below are my proofs.
Any helps or comments would be appreciated!

Proposition $1$: For all $x∈ \Bbb R$, $|x|^2 = x^2$.
Proof: Let $x∈ \Bbb R$ and suppose $x≥0$.
By the definition of the absolute value, $|x|$ is defined to be $x$ if $x≥0$.
Then $|x| = x$, and $|x|^2 = |x| \cdot |x| = x \cdot |x| = x \cdot x = x^2$.
Now, suppose $x<0.$ By the definition of the absolute value, $|x|$ is defined to be $–x$ if $x<0$.
Then $|x| = -x$ and $|x|^2 = |x| \cdot |x| = -x \cdot |x| = -x \cdot -x = x^2$.
Hence $|x|^2 = x^2$.

Proposition $2$: Let $x,y∈ \Bbb R$. Then $|x|<|y|$ if and only if $x^2 < y^2$.
Proof:
Let $x,y ∈  \Bbb R$ and suppose $|x|<|y|$. Then $|x| \cdot |x| < |x| \cdot |y| < |y| \cdot |y|$ and therefore, $|x|^2 < |y|^2$.
Then, by proposition $1$, $|x|^2 < |y|^2 \implies x^2 < y^2$.
Now, suppose $x^2 < y^2$.
Then, by proposition $1$, $x^2 < y^2 \implies |x|^2 < |y|^2$ and since $|x|,|y|≥0$,  $|x| < |y|$.
 A: You're pretty much right.
I would say your phrasing for the second proof could be a bit better to show, properly, each implication. Here's how I'd approach things...
Assume $|x|<|y|$. Then
\begin{align*}
x^2 &= |x|^2 \tag{proposition 1} \\
&= |x||x| \tag{def. of square} \\
&< |y||y| \tag{$|y|>|x|\ge 0$} \\
&=|y|^2 \tag{def. of square} \\
&=y^2 \tag{proposition 1}
\end{align*}
Thus $|x|<|y| \implies x^2 < y^2$. Now, conversely, suppose $x^2 < y^2$. Then
\begin{align*}
|x|^2
&= x^2 \tag{proposition 1} \\
&< y^2 \tag{assumption} \\
&= |y|^2 \tag{proposition 1}
\end{align*}
Thus, $x^2 < y^2 \implies |x|^2 < |y|^2$. Since $\sqrt z$ is monotone increasing on $\Bbb R_{\ge 0}$, it follows that $|x|^2 < |y|^2 \implies |x| < |y|$, the desired conclusion.
Alternatively, go for contrapositive: show that $|x| \ge |y| \implies |x|^2 \ge |y|^2$. I think this might be preferable to avoid invoking things that might not be "kosher" for the material at the time, however obvious. This proof is pretty straightforward:
\begin{align*}
|x|^2
&= |x||x| \tag{def. of square} \\
&< |y||y| \tag{$|y| \ge |x| \ge 0$} \\
&= |y|^2 \tag{def. of square}
\end{align*}
Thus, $x^2 < y^2 \implies |x|^2 <|y|^2 \implies |x|<|y|$, completing the proof.
That said, overall, I think you have the spirit of that proof right, and your first proof seems pretty good, so good job.

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.
