Prove that $|-x| = |x|$ Using only the definition of Absolute Value:
$\left|x\right| =  \begin{cases} x & x> 0 \\
-x & x < 0 \\
0 & x = 0,\end{cases}$
Prove that $|-x| = |x|.$
This seems so simple, but I keep getting hung up. I use the definition insert $-x$ into the definition, but I end up with:
$\left|-x\right| =  \begin{cases} -x & x> 0 \\
-(-x) & x < 0 \\
0 & x = 0\end{cases}$
which doesn't make sense to me. It certainly doesn't equal $|x|,$ does it?
I would use $|x| = \sqrt{x^2}$ but I am supposed to prove that identity later in the problem set. What am I doing wrong?

Your substitutions show that $|-x|=-|x|$. which is false.
The error comes from the fact that when introducing 
the minus sign you did not swap the sense of inequalities
once swapped
$|-x|= -x\quad  (x<0)$
$|-x| = x\quad  (x>0)$
$|-x| = 0\quad  (x=0)$
you see that $|-x|$ definition matches that of $|x|$ again.
