Because S_I is defined in the paper as the set of all total recursive functions compatible with the observations I. You cannot choose it to be anything else.

As I said, I am claiming to have a counterexample. I can choose any example out of all possible examples. You must show that the example I've chosen is not a possible case, to say that I'm not free to choose it.

S is not recursive, nor is it even recursively enumerable. That is, there is no Turing machine which, started on a blank tape, will output exactly all the members of S and no others. The same is true of S_I.

S is a subset of R, and R is the set of partial mu-recursive functions, meaning it consists of computable functions, meaning it is countable. R is enumerated by Goedel numbers; therefore it is enumerable. The elements of S_I are exactly those X for which p(X) > 0; and as I showed, they must be countable for this to occur. These are three separate proofs that S_I is countable in cardinality.

Constructing a Turing machine that generates S_I would be difficult. We don't have the information needed to do that. But saying that you don't know how to write that Turing machine, is not a proof that it does not exist.

S_I is of course countable, but there is no algorithm to calculate "the sum, over all possible worlds f in SI, of p(f)U(f(k))", because there is no way of enumerating SI, i.e. no computable bijection between S_I and N.

I've just demonstrated an example in which I provided a bijection between S_I and N, and computed the sum.