lo.logic - Proof of Gödel incompleteness

OK, if there's a model M of ZF which can be embedded in another model N via m, we cannot conclude that N thinks m is a model.

However, it seems to me the same proof can be carried without it, only using the following reverse implication for the infinite set of axioms ZF:

(*) if N is a model of ZF and m is any inside model of ZF, it can be lifted to an 'outside' model M, which moreover satisfies A iff N thinks m does

A is the diagonal sentence, saying there's a model which does not satisfy A

Now we can prove:
(1) if ZF is consistent, then A.

Proof: by the completeness Theorem ZF has a model M. if M is negative we're done. otherwise M has a negative sub-model which can be lifted by (*) to a negative model in the 'real world'. so again A is true.

(2) ( the inside version of (1) - for any theorem we can prove it has a valid proof - just write it down explicitly)
ZF proves that : if ZF is consistent, then A.

Now, if ZF is consistent, and it proves its consistency, then using modus ponens on (1) and 'inside' modus ponens of (2) we get:
(1') A
(2') ZF proves A. (Thus, any model satisfies A)

which is a contradiction: A is true iff there's a model that does not satisfy A.

Am I still missing something?