TITLE: Proof: $\mathbb Q \cap [0,1]$ is not compact (with the definition)
QUESTION [6 upvotes]: I have to prove that $\mathbb Q \cap [0,1]$ is not compact, directly with the definition with open covers (I am not allowed to use theorems like Heine-Borel).
My attempt: 
So I need to find an open cover of $\mathbb Q \cap [0,1]$ that has no finite subcover. I assume that we need to this by something like approximating an irrational number, i.e.
$$
U_n :=  \left]-1, \frac{\sqrt2}{2} - \frac{1}{n}\right[ \cup \left]\frac{\sqrt2}{2}+\frac{1}{n},2\right[
$$
This is a open cover of $\mathbb Q \cap [0,1]$ but I am not sure if this does not have a finite sub cover, like for example just $\mathopen]-1,2\mathclose[$
Any hints?

REPLY [3 votes]: More generally (and easily), consider any irrational number $\alpha\in[0,1]$. There is a decreasing sequence $(b_n)$ in $\mathbb{R}_{>0}$ (even in $\mathbb{Q}$, actually) such that $\alpha+b_0<1$ and $\lim_{n\to\infty}b_n=0$.
Then you can consider the open sets
$$
U_n=(-1,\alpha)\cup(\alpha+b_n,2)
$$
Clearly, $U_n\subseteq U_{n+1}$, for all $n$, and so
$$
U_{n_1}\cup U_{n_2}\cup \dots\cup U_{n_k}=U_m
$$
where $m=\max\{n_1,n_2,\dots n_k\}$. Since $\alpha<\alpha+b_m$, there is a rational number $q\in\mathbb{Q}\cap[0,1]$ with $\alpha<q<\alpha+b_m$. Therefore $\{U_{n_1}, U_{n_2},\dots, U_{n_k}\}$ cannot be a cover of $\mathbb{Q}\cap[0,1]$.
On the other hand, $\{U_n\}$ is an open cover for $\mathbb{Q}\cap[0,1]$. Indeed, let $q\in\mathbb{Q}\cap[0,1]$. If $q<\alpha$, $q$ belongs to every $U_n$. If $q>\alpha$, there exists $n$ such that $q>\alpha+b_n$, so $q\in U_n$.
Note that compactness means every open cover has a finite subcover; the finite covers do, but this is irrelevant.