How to prove that the Nested Interval Theorem fails to hold in $\mathbb Q$? Claim: The Nested Interval Theorem does not hold in $\mathbb Q$.
I can prove this by using sequences $a_n$ and $b_n$ where $a_n < b_n$ and they both converge for an $x$ which is any irrational number. In that scenario the intersection of such intervals would be empty. However, I don't find that good enough.
I've been thinking about following but somehow I don't find it quite right either.
Suppose The Theorem holds in $\mathbb Q$. Let $F_n=[a_n,b_n] \subset\mathbb{Q}$ where $a_n < b_n$. Because every closed set that is a subset of $\mathbb{Q}$ has maximum and minimum, they also have infimum and supremum. Therefore, let $M=sup\{a_n\mid n\in\mathbb{N}\}$ and $m=inf\{q_n\mid n\in\mathbb{N}\}$. Now $[M,m] \subset\ F_n$, $M<m$ and $M,m \in\mathbb{Q}$. Because $\mathbb{Q}\subset\mathbb{R}$, there is an $s\in\mathbb{Q}$ so that $M<s<m$. (It is assumed to be known that there is a rational number between every real number.) Now we have an interval $[M,s] \subset\ [M,m]$ which conflicts with $m$ being the infimum of upper bounds (because $s<m$ and $s\in\{b_n\mid n\in\mathbb{N}\}$. Therefore The Theorem can not hold in $\mathbb{Q}$.
I'm hoping for comments whether this proof is firm and if not, how should I do it.
 A: Here is a proof that does not make any reference to $\mathbb R$, and so may be "philosophically" more satisfying.
Towards a contradiction, assume that the Nested Interval Theorem holds in $\mathbb Q$. 
Since $\mathbb Q$ is countably infinite, one can enumerate it as a sequence $\{ r_n;\; n\in\mathbb N\}$, where the $r_n$ are pairwise distinct. Choose any rational closed interval $I_1$ having length $1$ and such that $r_1\not\in I_1$. Divide $I_1$ into 3 (rational) intervals of the same length. One of these intervals does not contain $r_2$; call it $I_2$. Then $I_2$ has length $3^{-1}$. Apply the same reasoning to $I_2$, and so on. This produces a decreasing sequence of closed intervals $(I_n)$ such that $I_n$ has length $3^{-n+1}$ for all $n$ and $r_n\not\in I_n$. Since the Nested Interval Theorem is assumed to hold in $\mathbb Q$, there is a rational number $r$ in the intersection of the intervals $I_n$. But by the very definition of the $I_n$, this number $r$ cannot be equal to any $r_n$; which is a contradiction.
A: Why are you dissatisfied with your first argument? It does the job just fine. Your second argument is wrong: the set $\{q\in\Bbb Q:\sqrt2<q<\sqrt3\}$ is a closed, bounded interval in $\Bbb Q$ with neither a maximum nor a minimum.
