Let $f$ be a function that maps an interval $[a, b]$ to some irrational number $r \in [a, b]$ that roughly splits the interval in half (e.g. both sides have at least 1/3 of the mass). Suppose we then index each $q \in \mathbb{Q} \cap [0, 1]$ with an infinite bitstring as follows: starting with the interval $i = [0, 1]$, write a $0$ if $q < f(i)$ or a $1$ if $q > f(i)$, and then recurse on the relevant subinterval to obtain the next bit, and so on.
This clearly gives an injection $g : \mathbb{Q} \cap [0, 1] \to \mathbb{B}^{\infty}$. My question is: what is $$\sup_n \text{ for every prefix $p$ of length $n$, there is a bitstring with prefix $p$ in $\mathop{range}(g)$}?$$ Clearly it is $\infty$ since any finite number is too small, but also $\aleph_0$ is "too big'' since this would imply that $\mathbb{Q} \cap [0, 1]$ is uncountable. How should I understand this? When we write $\sup_n = \infty$, is this a fundamentally different use of the symbol $\infty$ that doesn't correspond to an ordinal?
I am somewhat worried that I'm running into Axiom-of-Choice weirdness in thinking about this, since AC is needed to define $f$.
To say that the supremum in your question is $\infty$ means that $\infty$ is the least upper bound of the lengths $n$ described there. It does not mean that $\infty$ itself is one of those lengths. (It's the same idea assaying that $1$ is the supremum of the open interval $(0,1)$ doesn't mean that $1$ is itself a member of that interval.)