TITLE: What's wrong with this "proof" that Gödel's first incompleteness theorem is wrong?
QUESTION [18 upvotes]: Edit: I've added an answer myself, based on the other answers and comments.

Here is a very very informal "proof" (sketch) that Gödel's theorem is wrong (or at least that the idea of the proof is wrong) : 
Roughly, the proof of Gödel's theorem is as follows: For any decidable and consistent set of axioms $\Phi$ that contain (or imply) the first order Peano's axioms in first order language, we can construct a Gödel sentence $G^\Phi$, such that neither $\Phi\vdash G^\Phi$ nor $\Phi\vdash \neg G^\Phi$, but where we know from an argument in the meta-language that $G^\Phi$ is true. For any such $\Phi$, we will therefore have a counterexample to the completeness of the theory $Th(\Phi)$. Therefore we know that no such $\Phi$ can be complete (where complete means that all first order statements can be proven or disproven from it). 
Here is a failed proposal to "circumvent" Gödel's theorem that I have already heard someone make: Just define $\Phi_1=\Phi\cup \{G^\Phi\}$, and since we know that $G^\Phi$ is true in arithmetic, we know that $\Phi_1$ is consistent. The problem of course is: We can now formulate a new Gödel sentence $G^{\Phi_1}$, which cannot be proven from in $\Phi_1$, even though it is true in standard arithmetic. 

Now here is my proposal:
Rather than trying to add individual Gödel sentences to the set of axioms, we simply take the enumeration procedure, such that it enumerates $\phi_i \in \Phi$ for the original set of axioms $\Phi$, and also enumerates all successive Gödel sentences $G^{\Phi}, G^{\Phi_1}, G^{\Phi_2},...$. This is possible, since $\Phi$ is decidable, and decidable sets of finite strings are enumerable, so we can enumerate them successively, as $\phi_1, \phi_2, \phi_3$, where $\phi_1$ is the first statement of the enumeration of $\Phi$, and $\phi_2 = G^\Phi$, and $\phi_3$ is the second statement of the enumeration of $\Phi$, etc...
We can then define the set of axioms $\Phi_\infty = \{\phi_1, \phi_2, ...\}$. This will also have a Gödel sentence $G^{\Phi_\infty}$. But what we can simply do, is add this to the enumeration procedure as well. And then the next one, and the next, and so forth. We take this process to infinity, just as we did for $\Phi$, and just keep going. 
Every time a Gödel sentence pops up, we simply add it to the enumeration. Now note that: since the set of first order sentences is countable, the set of Gödel sentences is countable as well (since it is a subset of the set of first order sentences). Therefore we can in this procedure described above enumerate all possible Gödel sentences. 
The resulting set of sentences forms an enumerable and consistent set of sentences $\Psi$ that contains the original $\Phi$, and additionally 
contains the Gödel sentences of all possible sets of axioms $\Phi_x$. Therefore The Gödel sentence of $\Psi$ must be in $\Psi$ itself. 
Moreover, we can then create a "decidable version" of $\Psi$, by defining $\Psi^*=\{\psi_1, \psi_1 \land \psi_2, \psi_1 \land \psi_2\land \psi_3, ... \}$, for all $\psi_1, \psi_2,... \in \Psi$. We therefore have a consistent and decidable set of first order sentences that are true in standard arithmetic, contain  Peano's axioms, and bypass Gödel's proof of incompleteness.
This is obviously a contradiction with Gödel's theorem. So where is my "proof sketch" wrong?

REPLY [2 votes]: Some belated comments on your self-written answer:
This is all more or less correct. There's a theory referred to as "True Arithmetic", which is the set of all sentences in the language of arithmetic which are true in "the intended model".⁽¹⁾ By definition, True Arithmetic is consistent and complete, so by the first incompleteness theorem it must not be recursively enumerable. It follows that, for any theory of arithmetic T which is recursively enumerable, the set of sentences of True Arithmetic that T cannot prove is itself not recursively enumerable, as you observed.
I do have a nitpick with one thing you wrote, though it's not essential to the reasoning:

By " in principle" I mean that if an oracle told us the next godel
  sentence of this enumeration every time we asked for it, we could,
  using this oracle, enumerate all the godel sentences in my
  construction, and thereby "bypass" the limitations implied by godel's theorem.

This is all fine,⁽²⁾ as long as we keep in mind that the "next" Gödel sentence in the enumeration will not correspond with with the "next" theory as ordered by the corresponding ordinals. We do not in fact need an oracle to tell us the next sentence in the latter sense: the oracle is needed to give a single uniform listing of all the Gödel sentences, but the order in which it produces that list cannot correspond with the ordering of the corresponding ordinals.
To illustrate this, recall that your original idea depended on (recursive) lists of ordinals going some way beyond the finite ones. Such lists do indeed exist, but the order of the list cannot correspond to the natural order of the ordinals, since otherwise we'd never get past the finite ones. For example, here's a recursive list of all ordinals below ω+ω (where ω is the first infinite ordinal): 0, ω, 1, ω+1, 2, ω+2, ...
⁽¹⁾ The reference to "the intended model" is needed in order to accommodate the following fact: for any recursively enumerable theory of arithmetic T, there are models of T, i.e. sets of "numbers" together with definitions of 0, 1, +, and · which satisfy the axioms of T, and which are not "the intended model" because e.g. a given Gödel sentence is false in those models. This fact is a result of applying Gödel's completeness theorem (not to be confused with his incompleteness theorems!) to the existence a Gödel sentence.
Technically, the well-definedness of "the intended model", and hence of True Arithmetic, depends on the assumption that all sentences in the language of arithmetic do in fact have a "correct" truth value - an assumption that most working mathematicians take for granted, but that some people interested in the foundations of mathematics reject or at least question: see for instance http://jdh.hamkins.org/question-for-the-math-oracle/
⁽²⁾ If by "bypassing Gödel's theorem" you mean that the list of axioms obtained at the end of this process should be complete, i.e. sufficient for deducing all sentences of True Arithmetic, you might need to be careful about not only which Gödel sentences you list, but what order you list them in: this is the subject of ordinal notations, which I've mostly avoided except to note that the Church-Kleene ordinal, which we're effectively using here, doesn't have a recursively enumerable one.
There is more detail on this in the answers to https://mathoverflow.net/questions/67214/pi1-sentence-independent-of-zf-zfconzf-zfconzfconzfconzf-etc - a question close in spirit to yours, though it uses ZFC instead of Peano arithmetic, and the second incompleteness theorem rather than the first.