Absolute value problem $|x-y|=|y-x|$

$$|x-y|=|x-y|$$ $$|x-y|=|1|\cdot|x-y|$$ $$|x-y|=|-1|\cdot|x-y|$$ $$|x-y|=|-1\cdot(x-y)|$$ $$|x-y|=|y-x|$$

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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

$$|x-y| = |-(x-y)| =|-x+y| = |y-x|. $$