Is the set $E$ is open, closed, compact in $\mathbb{Q}$ 
Question: regard $\mathbb{Q}$ as a metric space with $d(p,q)=|p-q|$.
  Let $E$ be the set of all $p\in \mathbb{Q}$ such that $2<p^2<3$. Show
  that $E$ is closed and bounded, but $E$ is not compact. Also show that
  $E$ is open.

My attempt: given set $E$ is, 
$$E=\{p\in \mathbb{Q} : 2<p^2<3\}$$
Clearly, $E=(\sqrt 2, \sqrt 3)\cap \mathbb{Q}$ where $(\sqrt 2, \sqrt 3)$ is open set in $\mathbb{R}$. Hence $E$ is open in $\mathbb{Q}$. 
Similarly $E$ is closed because $E=[\sqrt 2, \sqrt 3]\cap \mathbb{Q}$.
$E$ is bounded is obvious. 
I don't know ** how to show $E$ is not compact in $\mathbb{Q}$ **. I know I have to find an open cover of $E$ which do not have finite subcover. But, is there is any other easy way to show $E$ is not compact in $\mathbb{Q}$? 
Further, is my solution for showing $E$ is open, closed in $\mathbb{Q}$ is correct? 
Please help.
 A: Proof of non compacity using sequential compactness
Consider the sequence defined by induction: $u_{n+1} = \frac{1}{2}\left(u_n+\frac{2}{u_n}\right)$ with $u_1=\frac{3}{2}$.
It is a sequence of rational numbers, decreasing and converging to $\sqrt{2}$ in $\mathbb R$. Hence $u_n \in E$ for all $n \in \mathbb N$.
$E$ is sequentially compact space as it is a metric space. As $(u_n)$ is a bounded sequence, it would have a converging subsequence if $E$ was compact. But that can't be as the limit is unique. So the limit would be $\sqrt{2}$ which doesn't belong to $E$.
Regarding your proof for openness, you missed the fact that $E$ has a negative part. So you should say
$$E=((-\sqrt 3, -\sqrt 2) \cap \mathbb Q) \cup ((\sqrt 2, \sqrt 3) \cap \mathbb Q).$$ And the union of open subspaces is open. Same for closeness as a finite union of closed subspaces is closed.
A: Consider the open cover $\mathcal U = \{(a,b) \subseteq \mathbf Q : \sqrt 2 < a < b<\sqrt 3\}$ of $E$. Suppose that $\mathcal U'$ is a finite sub cover of $\mathcal U$. There is some interval $(a,b)\in \mathcal U'$, such that for any other interval $(a',b') \in \mathcal U$: $a\leq a'$. Now note that there is a rational $u$ between $\sqrt  2$ and $a$. Then $u\in E$, but there is no interval in $\mathcal U'$ that contains $u$. A contradiction, hence $E$ is not compact.
A: *Mistake: You have $E= \{\sqrt 2\,,\,\sqrt 3)\cap \Bbb Q$ which is wrong. $$E=\{p\in \Bbb Q: 2<p^2<3\}=[\,(-\sqrt 3\,,\,-\sqrt 2) \cup (\sqrt 2\,,\,\sqrt 3)\,]\cap \Bbb Q.$$
**If $T$ is a topology on a set $X$ and $Y\subset X$ then the subspace topology $T_Y$ on $Y$ is $T_Y=\{t\cap Y: t\in T\}.$ And if $S\subset Y$ then $Cl_{T_Y}(S),$ the closure of $S$ in the space $Y,$ is $Y\cap Cl_T(S),$ where $Cl_T(S)$ is the closure of $Y$ in the space $X.$ In particular, $S$ is closed in $Y$ iff $S=Y\cap Cl_T(S).$
An important property, if  $T$ is generated by a metric $d$ on $X,$ is that  $T_Y$ co-incides with the topology on $Y$ generated by "the restriction of $d$ to $Y$". Applying this with $X=\Bbb R$ and $d(x,y)=|x-y|$ for all $x,y \in \Bbb R,$ and with $Y=\Bbb Q$ and $S=E,$ we have at once that $E$ is open and closed in $\Bbb Q,$ just as you said.
*** Let $$C=\{(-(\infty,\sqrt 2)\cap E\} \cup D$$ where $D= \{(y,\infty)\cap E:\sqrt 2\,<y\in \Bbb R\}.$
Now $C$ is an open cover of $E$ in the space $\Bbb Q.$
If $F\subset C$ then clearly $F$ is not a cover of $E$ if $F\cap D\ne \emptyset.$
But if $F$ is a $finite$ subset of $C$ and $F\cap D\ne \emptyset$ then there exists $y_0=\min \{y:(y,\infty)\cap E\in F\}$ and we have $\sqrt 2\,<y_0,$ so $\cup F$ is disjoint from the non-empty set $(\sqrt 2\,,y_0)\cap \Bbb Q =(\sqrt 2\,,y_0)\cap E,$ so $F$ is not a cover of $E$.
