This is an exercise from Terence Tao's analysis 1 book, chapter of integers and rational numbers.
Let $x$, $y$ and $w$ be rational numbers, and let $d(a,b):=|a-b|$ for any rationals $a$ and $b$.
If $d(y,x)\leq \epsilon$, $d(z,x) \leq \epsilon$ and $y\leq w \leq z$ or $z\leq w \leq y$, i.e, $w$ is a number between $z$ and $y$, prove that $d(w,x) \leq \epsilon$.
My attempt:
Since $|a-b|\leq |a|+|b|$, we have $d(y,x)=|y-x|\leq |y|+|x| \leq \epsilon$ and $d(z,x)=|z-x|\leq |z|+|x| \leq \epsilon$. Considering the case where $y\leq w \leq z$, we have $y-x\leq w-x \leq z-x$.
Now comes the problem, if the absolute value $| | $ operation could preserve order, so the proof would be done (because $|w-x|\leq|z-x| \implies d(w,x) \leq d(z,x) \leq \epsilon$). But this it's not generally true, so what can I do here?
Hint: use that $$|y-x| = d(y,x)\le\epsilon\iff-\epsilon\le y-x\le\epsilon.$$