Closed, Open, Bounded, but not Compact (Proof Verification) I'm practicing my proof-writing and was hoping you could let me know if this proof looks good. I would like to know if the proof is incorrect, if there are parts that are overly wordy/complicated, or if I'm missing some element of a proof that is helpful to see, if not strictly necessary.
The Prompt:
Regard $\Bbb Q$ as a metric space, with $d(p, q) =\vert p-q\vert$. Let $E:=\{p\in\Bbb Q\mid 2\lt p^2\lt 3\}.$
Show that $E$ is closed and bounded in $\Bbb Q$, but not compact. Is $E$ open in $\Bbb Q$?
My Proof:
Define sets $B$,$C$,$D$ such that:
$B = \{p \in\Bbb Q\mid\;p^2 < 2\}$
$C = \{p \in\Bbb Q\mid\;p^2 > 3\}$
$D = \{p \in\Bbb Q\mid\;p^2 = 2\;\lor\;p^2 = 3\}$
Note that $B \cup C \cup D \cup E =\Bbb Q$. First, we notice that any $e \in E$ is a limit (and interior) point of E, as there exists an open ball in $\Bbb Q$ centered at $e$ with radius $r =\min\{p-\sqrt{2},\sqrt{3}-p\}$ s. t. $B_r(e) \subset E$, so every neighborhood of $e$ contains elements of $E$ that are not $e$. No $b\in B$ is a limit point of $E$, as there is an open ball in $\Bbb Q$ centered at $b$ with radius $r=\sqrt{2}-p$ s. t. $B_r(b) \subset E^c$, so not every neighborhood of $b$ contains elements of $E$ that are not itself. The same holds for $C$ (with $r=p-\sqrt{3}$). Finally, we notice that $D=\emptyset$, as $\nexists p\in\Bbb Q$ s. t. $p^2 = 2\;\lor\;p^2=3$.  Therefore, every element of $E$ is a limit point of $E$, and no element of $B$,$C$, or $D$ is, so $E$ contains all its limit points, and is thus closed in $\Bbb Q$.
$E \subset (-3,3)$ so $E$ is bounded in $\Bbb Q$.
Let $\{G_n\}$ be a collection of open sets s. t. $G_n=\{p \in Q | 2 < p^2 < 3 - \frac{1}{n},\; n\in\Bbb N\}$. We can see that $E \subset \{G_n\}$, so $\{G_n\}$ is an open cover of $E$. However, there is no finite collection of sets in $\{G_n\}$ that covers $E$, so $E$ is not compact.
We saw earlier that every $e \in E$ is an interior point of $E$, so $E$ is an open set in $\Bbb Q$.
 A: We are given the metric space $\Bbb Q$ with the standard induced metric from $\Bbb R$. 
The following intervals are all open sets in $\Bbb R$:
$\quad (-\infty, - \sqrt 2)$
$\quad (-\infty, + \sqrt 2)$
$\quad (-\infty, - \sqrt 3)$
$\quad (-\infty, + \sqrt 3)$
$\quad (-\sqrt 2, +\infty)$
$\quad (+\sqrt 2, +\infty)$
$\quad (-\sqrt 3, +\infty)$
$\quad (+\sqrt 3, +\infty)$
We can write 
$\tag 1 B = [(-\infty, + \sqrt 2) \cap (-\sqrt 2, +\infty)] \, \bigcap \, \Bbb Q$
so $B$ is open.
Exercise 1: Show that $C$ and $E$ are also open in $\Bbb Q$.
Exercise 2: Show that the sets $B$, $C$ and $E$ forms a partition of $\Bbb Q$ into 3 blocks.
Exercise 3: Show that the sets $B$, $C$ and $E$ are all closed.
Hint a: The union of two open sets is open.
Hint b: The complement of an open set is closed.
Exercise 4: Show that the open ball $B_0(\sqrt 3)$ in $\Bbb Q$ satisfies the equation
$\tag 2 B_0(\sqrt 3) = [(-\infty, + \sqrt 3) \cap (-\sqrt 3, +\infty)] \, \bigcap \, \Bbb Q$
Exercise 5: Show that $E$ is contained in $B_0(\sqrt 3)$ and is therefore bounded (the distance between any two points in $E$ is less than $2 \sqrt3$).
Exercise 6: For every finite subfamily $\{G_m\}$ of the family of open sets $\{G_n\}$ there exist a $n_0$
such that
$\tag 3 \displaystyle \bigcup \, G_m = G_{n_0}$
Exercise 7: For each $k$, $G_k$ is a proper subset of $E$.
Hint: Between any two real numbers there exist a rational number.
Exercise 8: Using exercises 7/8 explain why $E$ is not a compact space.
