This is what I did, but I'm not sure if it's a good enough proof:
Since $|x|$ is equal to $x$ when $x$ is greater than or equal to 0, and is equal to $-x$ when $x$ is less than 0, I said that $|x|^2$ is equal to $(x)^2 = x^2$ when $x^2$ is greater than or equal to 0, and is equal to $(-x)^2 = x^2$ when $x^2$ is less than 0, therefore making $|x|^2$ = $x^2$. Is this valid?
Yes, this is absolutely valid! Sometimes, splitting the thoughts in two cases, $x<0$ and $x\geq 0$, is the way to go, since the very definition of $|x|$ depends on the sign of $x$, as you said.