Let $r\in \bf E$ , $r = \dfrac m n $.
Then we need to prove that there always exists a $\dfrac{p}{q}$ such that
$ \frac{m}{n}<\frac p q < \sqrt 3$
By exploting the special properties of $\sqrt 3 $,we can do it:
We start with
$\eqalign{ & r < \sqrt 3 \cr & r + 1 < \sqrt 3 + 1 \cr & \frac{1}{{r + 1}} > \frac{1}{{\sqrt 3 + 1}} \cr & \frac{1}{{r + 1}} > \frac{{\sqrt 3 - 1}}{2} \cr & \frac{{r + 3}}{{r + 1}} > \sqrt 3 \cr} $
But now we've gotten a number greater than $\sqrt 3 $. So what we'll do is apply the process again, and the relation will be reversed:
$\frac{{Q + 3}}{{Q + 1}} < \sqrt 3 $
Letting ${Q = \frac{{r + 3}}{{r + 1}}}$ will give
$\frac{{2r + 3}}{{r + 2}} < \sqrt 3 $
All we need to do now is prove that
$r < \frac{{2r + 3}}{{r + 2}}$
But since $r+2>0$ we get that
$\eqalign{ & {r^2} + 2r < 2r + 3 \cr & {r^2} < 3 \cr} $
which is true by hypothesis. Thus, given any rational $r$, there exists another rational $q$ such that
$r
where $q = \frac{{2r + 3}}{{r + 2}}$
Just to make things clear, I'll add some extra information.
The recursion defined as $r_0=1$ $r_{n+1}=\frac{2 r_n+3}{r_n +2} $ converges monotonically (increasing) to $\sqrt 3 $. This in particular means that given any $\epsilon>0$ there is an $r \in \rm E$ such that $\sqrt 3 -\epsilon < r$ which means $\sqrt 3 $ is the supremum of the set.
This also plays the role of showing that $3$ is irrational.