Why would I not have a free choice of S_I when constructing a counterexample? Remember that a counterexample is one particular case, taken out of all possible cases that a theorem applies to.

The only reason I would not have a free choice of S_I, is if each application of the theorem is meant to apply to all possible choices of S_I. That would mean that the theorem is actually proving that, for any unbounded utility function U, there exists some set of possible worlds for which it does not converge. That would be a still-interesting, but much weaker result.

Gödel numbering is used to enumerate R, the set of all partial recursive functions. R includes S and S_I, but neither S nor S_I is enumerable (by unsolvability of the halting problem).

Sorry; I believe you are incorrect, for the two reasons given in my post. Think about it: Goedel numbers are integers; how could anything that is enumerated by Goedel numbers not be enumerable? S_I, S, and R are all enumerable. The original paper says that R is the set of partial mu-recursive functions, which means computable functions; and the number of computable functions is enumerable.

EDIT: As AlephNeil noted, I meant that R is countable. "Neither S nor S_I is enumerable (by unsolvability of the halting problem)" is not supported by any argument. Nor does whether it is enumerable or not affect anything in my original post.