Dgdrxrt

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 $q in mathbbQ cap [0, 1]$ 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 $g : mathbbQ cap [0, 1] to mathbbB^infty$. My question is: what is
$$sup_n text for every prefix $p$ of length $n$, there is a bitstring with prefix $p$ in $mathoprange(g)$?$$
Clearly it is $infty$ since any finite number is too small, but also $aleph_0$ is "too big'' since this would imply that $mathbbQ cap [0, 1]$ 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.)