My first attempt at this proof divided into 2 cases, one where $x^2$ is greater than or equal to 0, and another where $x^2$ is less than 0.
For the first case, I said that the definition of absolute value says that $|x|^2$ = $x^2$.
For the second case, I said that the definition of absolute value says that $|x|^2$ = $(-x)^2$ = $x^2$. I then concluded to say that since both cases were correct, the proof was completed.
This was marked incorrect by my instructor, and he marked that I was not considering $|x^2|$. I'm assuming by this that I used the absolute value definition to show that $|x^2|$ = $x^2$, which isn't the same thing as saying $|x|^2$ = $x^2$. Another thing he pointed out was that the case $x^2 < 0$ never happens, which I probably should have noticed the first time.
I feel like the proof should be shown by me showing that $x^2$ is always positive, but I don't know if I can outright say that without giving the formal definition of $x^2$. Another thing I wanted to try was to show that $|x|^2$ = $|x^2|$ through saying $|x|^2$ = $|x| * |x|$ and then using the theorem that $|xy|$ = $|x||y|$ to conclude that $|x| * |x|$ = $|x^2|$. But I think that would take me down the same route as last time, which would be marked incorrect.
The cases to examine are (i) $x\ge 0$ and (ii) $x\lt 0$.
In Case (i), we have $|x|^2=x^2$.
In Case (ii), we have $|x|^2=(-x)^2=x^2$. Conceivably one might want to insert an extra step, $(-x)^2=(-1)^2x^2=x^2$.