Prove that $|x|<|y| \iff x^2<y^2$.

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My proof is below. I am not sure about the second part.
Proof:
First I prove $|x|<|y| \implies x^2<y^2$.
Suppose $|x|<|y|$. Then,
\begin{align*} |x||x| < |y||x| &\implies x^2<|y||x|\tag1\\ |y||x| < |y||y| &\implies |y||x| < y^2\tag2 \end{align*} By $(1)$ and $(2)$, $x^2<y^2$
Next I prove $x^2<y^2 \implies |x|<|y|$. For that consider the contrapositive.
$$|x| \geq |y| \implies x^2 \geq y^2$$ From previous case $|x|>|y| \implies x^2>y^2$ is true. Therefore the contrapositive is true. Hence the result follows.

Is my proof correct ?