I think your comparison misses an important point. If you pick in every step a sentence that is non-provable from your current axioms, but not contradictory with them either, after infinitely many steps[1] you will have picked all of them, and get a system called True Arithmetic, in which every true sentence is provable, essentially by definition. The reason why Goedel's incompleteness theorem does not work here is that your set of axioms is not recursively enumerable, and this is a necessary assumption of GIT.
[1] - If "after infinitely many steps" sounds fishy to anyone, please keep in mind that you start with an infinite list of axioms anyway -- first order Peano arithmetic theory has an induction axiom schema Ind that basically says that for every sentence f, Ind(f) (the sentence you get by substituting every occurrence of a single free variable in Ind with f) is an axiom of PA.
The "infinitely many steps" does sound fishy to me, because I wonder whether the number of axioms in True Arithmetic is countably infinite or uncountably so, and whether it makes a difference (I think it does).
But yes, the comparison is an analogy, certainly not correct at every level.
The set of all statements is easily seen to be countable, so the subset of statements being true, being infinite, is countable as well.
It does not make any difference, though, since what matters here is, as I said, recursive enumerability of axioms. Anyway, transfinite induction lets one use "pick next element" arguments even on uncountable sets.
[1] - If "after infinitely many steps" sounds fishy to anyone, please keep in mind that you start with an infinite list of axioms anyway -- first order Peano arithmetic theory has an induction axiom schema Ind that basically says that for every sentence f, Ind(f) (the sentence you get by substituting every occurrence of a single free variable in Ind with f) is an axiom of PA.