Choosing a Cauchy sequence for a real

Such a choice mechanism would translate to a canonical choice for a function $\Bbb N \to \Bbb N_{>0}$ from a given set $S$ of such functions. If I'm not mistaken, this is some nontrivial choice principle (and if I am, please do correct me).

Given $f: \Bbb N \to \Bbb N_{>0}$, define: $$b_f (n) = \begin{cases}2^{-k} &: n = \sum\limits_{i=0}^k f(i) \\0&: \text{otherwise}\end{cases}$$

Now define $s_f(n) = \sum\limits_{i=0}^n b_f(i)$. Then for all $f$: $$\lim_{n\to\infty} s_f(n) = 2$$ and $f \leftrightarrow s_f$ is a bijective correspondence. Hence a choice function for each set $\{s_f: f \in S\}$ amounts to a choice function for each $S \subseteq (\Bbb N_{>0})^{\Bbb N}$.

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By defining, for $f: \Bbb N \to 2$: $$b_f(n) = \begin{cases}2^{-k} &: n = k + \sum\limits_{i=0}^k f(i) \\0&: \text{otherwise}\end{cases}$$ the argument seems to carry over to $2^{\Bbb N}$.

Let me take Lord_Farin's answer, and crank it all the way to $2^{\aleph_0}$.

Suppose that we could have chosen from any family of Cauchy sequences. Fix for each real number a canonical Cauchy sequence of rationals $r_n$ which is strictly increasing. This is of course doable without choice.

Now suppose that $A_r\subseteq 2^\omega\setminus 2^{<\omega}$ is non-empty for each $r\in\Bbb R$, then consider $C_r=\{\langle r_n\mid a_n\neq 0\rangle\mid\langle a_n\rangle\in A_r\}$. Namely we encode each $a\in A_r$ as a subsequence of $r_n$, with the assumption that $a$ is not eventually $0$.

By the assumption there will be a choice from each $C_r$, and then we can easily decode this into a choice from $A_r$. So we have proved the axiom of choice for families of size $\leq2^{\aleph_0}$ of sets of reals. And of course we cannot even prove choice for countable families of sets of reals in $\sf ZF$ itself.