Show that a subset $E$ $\subset \Bbb Q$ is not compact in $(\Bbb Q, d)$ and decide whether it is open or not 
Assume $(\Bbb Q, d),$ $d(p, q):= |p -q|$
is a metric space and
$E := \{p \in \Bbb Q : 2 < p^2 < 3\} = \{p \in \Bbb Q : \sqrt2 < p < \sqrt3\} \subset \Bbb Q.$
I have to show that $E$ is not compact, and furthermore, I have to decide whether E is open or not.
Note: I already proved that $E$ is closed and bounded since this was also part of the task.

Approach
Assume $E$ is compact. Then, for every open cover $(U_{i_k})_{i_k \in I}$, there must be a finite number of indices $i_1, ..., i_n \in I$ such that
$$A \subset U_{i_1} \cup ... \cup U_{i_n}.$$
We can assume that $U_{i_k} \subset E.$
Since $\sqrt2 \notin E$, there is no $p \in E$ such that
$$p = \sqrt2.$$
Therefore, and since $E \subset \Bbb Q$ must be countable infinite, we would always find an $\epsilon \gt 0$ such that
$|p - \sqrt2| \le \epsilon$ $\forall p \in E.$ (which already proves that $E$ is open when we show the same thing for $\sqrt3$, but this is obvious)
Since p "converges" to $\sqrt2$ (there might be a better way to express this?), it follows directly that the number of covers can not be finite such that $E$ can not be compact.
 A: To show non-compactness, we shall just formalise your convergence idea.  
Let $a_n$ be a strictly decreasing sequence tending to $\sqrt{2}$ (for example take alternating terms of the truncated continued fraction) and let $b_n$ be a strictly increasing sequence tending to $\sqrt{3}$. 
Consider the sets $U_n=(a_n,b_n)\cap \mathbb{Q}$ (open by the same argument you gave), then $\cup U_n = (\sqrt{2},\sqrt{3})\cap \mathbb{Q}$ and are increasing. Thus the union of any finite subset of $U_n$'s looks like $U_N$ where N is the maximum index within this subfamily. Now, as $\sqrt{2}$ is not rational, $U_N \neq E$ and so E is not compact.
A: Consider a set of the form $(a,b)\cap\mathbb{Q}$; it's useful to consider $\mathbb{Q}$ as a subspace of its completion $\mathbb{R}$), because this tells us that $(c,d)\cap\mathbb{Q}$ is open in $\mathbb{Q}$, for any choice of $c,d\in\bar{\mathbb{R}}=\mathbb{R}\cup\{-\infty,\infty\}$.
Suppose now $a$ is irrational. Then consider a decreasing sequence $(a_n)$ such that $a_n\in\mathbb{Q}$ and $\lim_{n\to\infty}a_n=a$.
Then the sets
$$
(a_n,b)\cap\mathbb{Q}
$$
provide an open cover of $(a,b)$ with no finite subcover.
