Why doesn't cantor's theorem work with nested intervals work with rationals? I understand how and why does the cantor's theorem with nested intervals proof work. I'm also aware that the nested interval property doesn't generally work for $\Bbb Q$ - rational numbers - but I still can't see why the proof can't be used for rational numbers.
My point is, while it is true that the nested interval property doesn't work generally for $\Bbb Q$, it should work (i.e. intersection of subintervals shouldn't be empty) when we guarantee that the borders of each (sub)interval are always rational, because then supremum and infinum of respective borders are within $\Bbb Q$. 
And the proof as outlined in the link provided above kind of implies that it should be possible to construct the interval in that way. The 'one-third' interval that doesn't feature $x_n$ can always be bordered with rational numbers (or can you prove me otherwise?) because the only two properties it needs to satisfy is to not to include $x_n$ and to be subinterval of the previous one. And if that's the case, then intersection of such interval(s) should always exist on $\Bbb Q$, or shouldn't it? And if it does, then the whole R proof should work the same for $\Bbb Q$ (which it can't, because $\Bbb Q$ are countable (and many other proves show that)).
PS 1: If it's possible to prove that that's not the case, that you can't construct the 'one-third' intervals in a way they have an intersection, then I completely understand why the proof doesn't work for $\Bbb Q$.
PS 2: I understand that generally the intersection of sub-intervals doesn't have a element in $\Bbb Q$. E.g. for $[\sqrt 2, \sqrt 2 + 1 / n ]$ but then again, this interval doesn't have rational boundaries.
 A: Emanuele Paolini's accepted answer in that question said

Suppose by contradiction that $x_1,x_2,\dots,x_n,\dots$ is an enumeration of $I_0=[0,1]$. Take the interval $[0,1]$ and consider the two thirds intervals: $[0,1/3]$, $[2/3,1]$. Let $I_1$ be one of these intervals which does not contain $x_1$. Then proceed by dividing $I_1$ in three parts and considering two of them which are disjoint. Let $I_2$ be the one not containing $x_2$. 
Going on like that you find a sequence of intervals: $I_1 \supset I_2 \supset \dots \supset I_n \supset \dots$.  By the continuity axiom there is a point $\bar x$ in the intersection of all these intervals. By construction this point is different from any $x_n$ since $\bar x \in I_n$ while $x_n \not \in I_n$.

Now create a sequence such that every time you chose $I_n$ to be the left-hand set $a_n=0$ while if $I_n$ to be the right-hand set $a_n=2$.  It turns out that single point of intersection would be $\bar x=\sum_n a_n 3^{-n}$, i.e. the tertiary fraction $0.a_1a_2a_3\cdots$.  If you were only intersecting subsets of the rationals then the infinite intersection would be empty. 
If all the $x_n$ are rational then this does not lead to a contradiction in the same way as with the reals as there are plenty of irrational tertiary fractions (i.e. not recurring) just using the digits $0$ and $2$. So you can conclude that the end of the process leads to a number $\bar x$ which is not rational.     
