In [this question](https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice) the following observation was made:

Consider a sequence of boxes numbered 0, 1, ... each containing one real number. The real number cannot be seen unless the box is opened. 

Define a *play* to be a series of *steps* followed by a *guess*. A *step* opens a set of boxes. A *guess* guesses the contents of an unopened box.  A *strategy* is a rule that determines the steps and guess in a play, where each step or guess depends only on the values of the previously opened boxes of that play. 

Then for every positive integer $k$, there is a set $S$ of $k$ strategies such that, for any sequence of (closed) boxes, there is at at most one strategy in $S$ that guesses incorrectly.

My question is this: Can $k$ be countably infinite (instead of a positive integer)? If not, is there a proof?

[Edit: the original question also asked whether $k$ can be uncountable; this was answered by Dan Turetsky in the negative in comments].

The best I have been able to show is that, if the function $f:\mathbb{N}\to\mathbb{R}$ defined by the contents of the initial sequence of boxes is recursive (viewing elements of $\mathbb{R}$ as binary sequences), then $k$ can be countably infinite. To see this, call a subset $X$ of $\mathbb{N}$ *signature* if two recursive functions on $\mathbb{N}$ that eventually agree on $X$ also eventually agree on $\mathbb{N}$. (Two functions "eventually agree" if they differ in finitely many places). Call two Turing Machines *equivalent* if their associated functions on $\mathbb{N}$ are equivalent (that is, eventually agree). A diagonalization argument on the class representatives of the Turing Machines yields an infinite partition $U$ of $\mathbb{N}$ into signature subsets. The $i$'th strategy in $S$ first opens all the boxes whose indices are not in the $i$'th element $U_i$ of $U$, determines the class representative Turing Machine T that generates the resulting values on the opened boxes for boxes whose indices are greater than $N$ (for some positive $N$), and guesses that a box with index greater than $N$ and in $U_i$ has a value specified by $T$. 

However, I have not been able to modify this for the non-computable case.