I think the best way is just to work it out with brute force.
Then, having $|x-y| + |y-z| = |x-z|$,assume $y$ is not between $z$ and $x$; without loss of generality (because the other cases will be proved analogously), we suppose $y < x$ and $y \leq z$.
Also, without loss of generality, we assume $x \leq z$ (as already pointed out in the comments above, we can just prove that $y$ is between $x$ and $z$, without assumption on who's greater between them).
Then, by the implication you've already proved, since $y < x \leq z$ we have
$|y-z|=|y-x|+|x-z|$; substituting in the original identity
$$|x-y| + |y-z| = |x-z| \iff |x-y| + |y-x|+|x-z| = |x-z| \iff 2|x-y|=0$$ contradiction.