Proof Verification: show that $[0,1] \cap \Bbb Q$ is not compact in the metric space $\Bbb Q$ Show that $[0,1] \cap \Bbb Q$ is not compact in the metric space $\Bbb Q$.
Pick $\alpha \in [0,1]$ such that $\alpha$ is irrational.
Let $U_0 = \{q \in \Bbb Q : q \lt \alpha\}$ and let $U_n = \{q \in \Bbb Q : q \gt \alpha + 1/n\}, \forall n \in \Bbb N$.
$U_0$ and $U_n$ are open in $\Bbb Q$ ($\forall n \in \Bbb N$) as they describe open intervals.
Pick $p \in [0,1] \cap \Bbb Q$. If $p \lt \alpha$, then $q \in U_0$. Otherwise, if $p \gt \alpha$, then $\exists n \in \Bbb N$ such that $p \gt \alpha + 1/n$, so $p \in U_n$ for some $n \in \Bbb N$. Thus, $[0,1] \cap \Bbb Q \subseteq \cup_{n=0}^{\infty}U_n$.
Hence, $\{U_n\}_{n=0}^{\infty}$ is an open cover of $[0,1] \cap \Bbb Q$.
Suppose for contradiction that $\{U_n\}_{n=0}^{\infty}$ has a finite subcover. Then $\exists n_1 \lt \ldots \lt n_k$ such that $[0,1] \cap \Bbb Q \subseteq \cup_{i=1}^k U_{n_i}$.
It is clear that we must have $n_1 = 0$, as otherwise $[0,\alpha] \cap \Bbb Q$ is not covered (notably, $0 \notin \cup_{i=1}^k U_{n_i}$ if $n_1 \neq 0$).
Then, notice that since $n_2 \lt \ldots \lt n_k$ and $U_n = \{q \in \Bbb Q : q \gt \alpha + 1/n\}$, we have $U_{n_2} \subseteq U_{n_3} \subseteq \ldots \subseteq U_{n_k}$. Thus $[0,1] \cap \Bbb Q \subseteq U_{n_1} \cup U_{n_k}$. By the density of rationals, since $\alpha \in [0,1]$, $\exists p \in [0,1] \cap \Bbb Q$ such that $\alpha \lt p \lt \alpha + 1/n_k$.
Then, $p \notin U_0$, since $p \gt \alpha$, and $p \notin U_{n_k}$, since $p \lt \alpha + 1/n_k$.
So, in fact, $U_{n_1} \cup U_{n_k}$ does not cover $[0,1] \cap \Bbb Q$, a contradiction.
Thus, $\{U_n\}_{n=0}^{\infty}$ has no finite subcover, so $[0,1] \cap \Bbb Q$ is not compact in $\Bbb Q$.
Any criticism or correction is appreciated!
 A: We know $\mathbb{Q}$ is dense in $\mathbb{R}$. So exist $\left\{ {{x_n}} \right\} \subset \mathbb{Q} \cap \left[ {0,1} \right]$ such that $\mathop {\lim }\limits_{n \to  + \infty } {x_n} = \frac{{\sqrt 2 }}{2}$. We will prove not exist subsequence of $\left\{ {{x_n}} \right\}$ by contraction. If exist subsequence of  $\left\{ {{x_n}} \right\}$ (we denote this subsequence is $\left\{ {{x_{{n_k}}}} \right\}$) such that $\mathop {\lim }\limits_{k \to  + \infty } {x_{{n_k}}} = \alpha  \in \mathbb{Q} \cap \left[ {0,1} \right]$. Because $\mathop {\lim }\limits_{n \to  + \infty } {x_n} = \frac{{\sqrt 2 }}{2}$ so $\alpha  = \frac{{\sqrt 2 }}{2}$. But $\frac{{\sqrt 2 }}{2} \notin \mathbb{Q} \cap \left[ {0,1} \right]$. So $\mathbb{Q} \cap \left[ {0,1} \right]$ is not compact in the metric space Q
A: Here is a similar approach that avoids some technicalities by being a disjoint open cover.
Choose irrational $\alpha_0$ such that $0 < \alpha_0 < \alpha_0+{1 \over 2} < 1$.
Now consider the points $\alpha_k = \alpha_0 + {1 \over 2^k}$, $k = 1,2,...$, and note that they are all irrational.
Now consider the open cover $[0,\alpha_0) \cap \mathbb{Q}, (\alpha_0+{1 \over 2},1]\cap \mathbb{Q}, (\alpha_{k+1}, \alpha_k)\cap \mathbb{Q}$, for $k =1,2,...$. It is clear that they form an open
partition of $[0,1] \cap \mathbb{Q}$.
Since they form a partition, any finite sub partition only covers a strict subset of
$[0,1] \cap \mathbb{Q}$.
