Specific instance of a general selection principle in topology. I am trying to figure out how to approach a proof of the following claim:
For any sequence $\langle U_n: n \in \mathbb{N} \rangle$ of bases for the standard topology on $\mathbb{Q}$, there is a sequence $\langle F_n: n \in \mathbb{N} \rangle$ such that each $F_n \in U_n$ and $\{F_n: n \in \mathbb{N}\}$ is a basis for the standard topology on $\mathbb{Q}$. 
So for a hint of how that proof might go, I wonder how one could show the following:
Let $(\mathbb{Q}, \tau)$ be the standard topology on the rationals. 
Define a sequence $\langle U_n: n \in \mathbb{N} \rangle$ as follows: $U_n = \{I:I \text{ an open interval of length } < \frac{1}{2^n}\}$. 
Can we find a sequence $\langle F_n: n \in \mathbb{N} \rangle$ such that for any $n$, $F_n \in U_n$ and $\{F_n: n \in \mathbb{N}\}$ is a basis for $(\mathbb{Q}, \tau)$?
 A: $\Bbb Q$ is countable, so you can enumerate $\Bbb Q=\{q_n:n\in\Bbb N\}$. Let $\mathscr{B}=\{B_n:n\in\Bbb N\}$ be any countable base for $\Bbb Q$. You have to make sure to choose your sets $F_n$ so that for $k,m\in\Bbb N$ there is an $n\in\Bbb N$ such that $q_k\subseteq F_n\subseteq B_m$. One way to do this is to make use of the pairing function $\pi:\Bbb N\times\Bbb N\to\Bbb N$; it’s a bijection, so it has an inverse bijection $\pi^{-1}:\Bbb N\to\Bbb N\times\Bbb N$. For each $n\in\Bbb N$ choose $F_n$ so that 


*

*$F_n\in U_n$, and  

*if $\pi^{-1}(n)=\langle k,m\rangle$ and $q_k\in B_m$, then $q_k\subseteq F_n\subseteq B_m$.


You have to prove that this is always possible, but that’s not too hard if you remember that each $U_n$ contains arbitrarily small open intervals.
And of course you have to prove that the resulting collection $\{F_n:n\in\Bbb N\}$ is a base for $\Bbb Q$, but the construction is set up to make that fairly straightforward.
A: Fix an enumeration $\{(q_n,I_n):n\in\mathbb N\}$ of all pairs of the form $(p,I)$ where $I$ is an open interval with rational end-points and $q\in I$. Now, for each $n$ pick $F_n\in U_n$ such that $q_n\in F_n\subseteq I_n$ this can be done since $U_n$ is a basis. It should be clear from the construction that $\{F_n:n\in\mathbb N\}$ is a basis.  
