Absolute value problem $|x-y|=|y-x|$
$$|x-y|=|x-y|$$ $$|x-y|=|1|\cdot|x-y|$$ $$|x-y|=|-1|\cdot|x-y|$$ $$|x-y|=|-1\cdot(x-y)|$$ $$|x-y|=|y-x|$$
Without $|x||y|=|xy|$
If $x>y$
Since $y-x<0$ that means $|y-x|=-(y-x)=x-y$
$$|y-x|=x-y$$
Since $x-y>0$ that means $|x-y|=x-y$
$$|x-y|=x-y$$
Equality is transitive
$$|x-y|=|y-x|$$
If $y>x$
Since $x-y<0$ that means $|x-y|=-(x-y)=y-x$
$$|x-y|=y-x$$
Since $y-x>0$ that means $|y-x|=y-x$
$$|y-x|=y-x$$
Equality is transitive
$$|x-y|=|y-x|$$
The case of $x=y$ is left as an exercise for the reader.
You say that you have proven that $|x|=|-x|$, then it immediately follows that
$$|x-y| = |-(x-y)| =|-x+y| = |y-x|. $$