Deriving a bounding $\delta$ of an interior point This question is based on the Baby Rudin's 2.16:

Regard $Q$, the set of all rational numbers, as a metric space, with $d(p,q)=\lvert p -q \rvert$. Let $E$ be the set of all $p \in Q$ such that $2 < p^2 < 3$. <...> Is E open in Q?


The solution manual defines $\delta$ as the minimum of $\sqrt{\frac{3-x^2}{3}}$, $\frac{3-x^2}{3\lvert x \rvert}$ and $\frac{x^2 - 2}{2 \lvert x \rvert}$. Then, if $2 < x^2 < 3$ and $y \in (x-\delta, x+\delta)$ we must have $2 < y^2 < 3$.
I'm trying to understand why my bounds are wrong and how they were obtained in the manual. As I understand, we need to show that every $p \in \{x \mid 2 < x^2 < 3 \}$ is an interior point, i.e. there is a bound which I defined as $\delta^\star := \min(\lvert p \rvert - \sqrt{2}, \sqrt{3} - \lvert p \rvert)$ so that we have $2 < q^2 < 3$ for every $q \in (p - \delta^\star, p+\delta^\star)$.
Thank you for any hints.
 A: I would suggest a different way of attacking this one:
Define the following, for $\epsilon>0$:
$B_\epsilon(x, \mathbb{R})= \left\{ y \in \mathbb{R}: |x-y|<\epsilon \right\}$
$B_\epsilon(x, \mathbb{Q})= \left\{y \in \mathbb{Q}: |x-y|< \epsilon \right\}$
It is then clear that $X \subseteq \mathbb{R}$ is open if and only if for each $x \in X$ there is an $\epsilon>0$ such that $B_\epsilon(x, \mathbb{R}) \subseteq X$. Similarly, $X \subseteq \mathbb{Q}$ is open if and only if for each $x \in X$ there is an $\epsilon>0$ such that $B_\epsilon(x, \mathbb{Q}) \subseteq X$ (both of these are by definition of the topology on a metric space). 
We are now in a position to prove a result which is actually more general:
Let $U \subseteq \mathbb{Q}$. If $U= \mathbb{Q} \cap V$ for some open $V \subseteq \mathbb{R}$, then $U$ is open in $\mathbb{Q}$. 
$\textbf{Proof:}$ Suppose $U= \mathbb{Q} \cap V$ for some open $V \subseteq \mathbb{R}$ and take $x \in U$. Then $x \in V$, so there is some $\epsilon>0$ such that $B_\epsilon(x, \mathbb{R}) \subseteq V$. Then $B_\epsilon(x, \mathbb{R}) \cap \mathbb{Q} \subseteq \mathbb{Q} \cap V= U$ and $B_\epsilon(x, \mathbb{R}) \cap \mathbb{Q}=B_\epsilon(x, \mathbb{Q})$, so there is an $\epsilon>0$ such that $B_\epsilon(x, \mathbb{Q}) \subseteq U$. Since $x \in U$ was arbitrary, $U$ is open in $\mathbb{Q}$. $\Box$
Now, we have $E= \mathbb{Q} \cap (\sqrt2, \sqrt3)$. $(\sqrt2, \sqrt3)$ is open in $\mathbb{R}$, so $E$ is open in $\mathbb{Q}$ by the above.  
