Exercise in Tao Analysis Book I'm currently studying in the book of Analysis of Terry Tao, amazing  book by the way. In one exercise I'm not pretty sure about how can do it (I know that will be almost trivial but I'm stuck in it).
Definition: Let $\varepsilon >0$ and $x,y\in \mathbb{Q}$. We say that $y \text{ is } \,\,\varepsilon\text{ - close to } x \iff d(x,y)\le \varepsilon$; where $ d(x,y)$ is the distance of $y$ to $x$ define as $d(x,y)= \,\mid x-y\mid.$
Ex: Let $\varepsilon>0$. If $x$ and $y$ are $\varepsilon \text{ - close}$ to $x$, and $w$ is between $y$ and $z$, then $w$ is also $\varepsilon \text{ - close}$ to $x$.
So my question: is there a clever way to do it without divide the exercise by cases when $y\le w\le z$ and then when $z\le w\le y$? Sorry if I'm not put my attempt, everything I tried is completely wrong. Thanks in advance. 
 A: One characterization/definition/property of "between" is that $d(x,w) + d(w,y) = d(x,y)$.  The distances to $x$ and $y$ are bounded by the $xy$ distance.
A: There is a typo in the original post, saying "if x and y are e-close to x", instead of "y and z" being e-close to x.
Claim: Let $\epsilon > 0$. If $y$ and $z$ are both $\epsilon$-close $x$, and $w$ is between $y$ and $z$ (i.e., $y \leq w\leq z$ or $z\leq w \leq y$), then $w$ is also $\epsilon$-close to $x$.
Pf. Suppose $y \leq w \leq z$. We need to show that $d(x, w) \leq \epsilon$. By definition it suffices to show that $|x - w| \leq \epsilon$, i.e. that, $-\epsilon \leq x - w \leq \epsilon$ or that $w \leq x + \epsilon$ and $x - \epsilon \leq w$. To prove the claim, we show that these last two inequalities hold.
Since $d(x, y) \leq \epsilon$, we have $-\epsilon \leq x - y \leq \epsilon$. From the last inequality we have that $x - \epsilon \leq y$, and since $y \leq w$, by transitivity, $x - \epsilon \leq w$, which shows that the second inequality holds.
On the other hand since $d(x, z) \leq \epsilon$, $-\epsilon \leq x - z \leq \epsilon$. It follows that $z \leq x + \epsilon$ and since $w \leq z$, by transitivity again, we conclude that $w \leq x + \epsilon$, which shows that the first inequality also holds and proves the claim for the case where $y \leq w \leq z$. The other case can be proved similarly.
