What's wrong with this proof that ZFC is consistent If we take ZFC minus all it's axioms, we can easily prove that ZFC's first axiom is independent (because all statements are independent in a formal system without any axioms). We can then prove that the second axiom is independent of the first one, that the second is independent of the first two, and so on, until we prove that ZFC is consistent.
Doesn't this contradict Godel's inconsistency theorem which says that the consistency of a formal system cannot be proved?
Also, by continuing with the method described above, and repeatedly assuming that the shortest independent statement is true, wouldn't we end up, after an infinite amount of repetitions, with a system that is both complete and consistent?
 A: No, you can't do that. The most basic reason why is that the question, "Is $\varphi$ independent of $T$?" is not decidable - given an arbitrary first-order sentence $\varphi$ and a finite set of first-order sentences $T$, there is no way to tell if $\varphi$ is independent of $T$. Specifically, the set of pairs $$\{(\varphi, T): \mbox{$\varphi$ is a sentence, $T$ is a finite set of sentences, $T\not\vdash\varphi$, and $T\not\vdash\neg\varphi$}\}$$ is (when appropriately coded as a set of natural numbers) not computably enumerable.
Concretely, this means that - fixing some background recursive consistent set of axioms $S$ extending (say) ZFC - there is some finite theory $T$, and some sentence $\varphi$, such that (1) $\varphi$ is independent of $T$, but (2) $S$ doesn't prove that. At some point in your process, the following will happen: you've listed sentences $\varphi_1, . . . \varphi_n$ that you're confident are consistent with each other, and you ask - "Is $\psi$ consistent with $\varphi_1\wedge . . .\wedge\varphi_n$?" Unfortunately, your theory can't answer this question, and so your process gets stuck.

Now, what about if you run your process without regard to computational complexity? In this case you can indeed define a consistent, complete theory $T$! The problem is, this $T$ isn't computable. So the existence of such a $T$ doesn't contradict Godel's theorem at all.
ADDED: You might well ask, "What do such $T$ look like?" Well, bad news first: without specifying the order in which you look at sentences, I can't tell you much.
Good, and very surprising, news: as long as the order you look at sentences in is computable, I can tell you some things! Specifically, I can tell you that $T$ is wrong: $T$ will have axioms which are false statements about the natural numbers. (Ignoring for the moment the philosophical questions around what I mean by "false" - for now, just assume we have something we agree is the "real" set of natural numbers.) This is because of computability theory: the theory $T$ will be $\Delta^0_2$, but the true theory of arithmetic is much more complicated than that. And we know this without understanding any of the details about $T$!
Neat, huh?

One more commment: ZFC has infinitely many axioms. Even if we could do this for each finite subset of ZFC, we would need to be able to prove - in ZFC - that we could do this for each finite subset of ZFC! But "$\forall$" and "$\vdash$" don't commute: just because, for each $n$, $ZFC$ proves $p_n$, doesn't mean $ZFC$ proves "$\forall n p_n$."
One example of this is that ZFC (assuming it's consistent of course :P) proves "There is no contradiction in the ZFC axioms of length $n$" for each $n$. But, by Godel, ZFC doesn't prove "For every $n$, there is no contradiction in the ZFC axioms of length $n$."
In fact - and this is a point where ZFC is special (by contrast, everything I've said above is true for every reasonable theory of set theory and arithmetic) - ZFC can do more! For each finite $F\subset ZFC$, ZFC proves F is consistent! This is called the reflection theorem (see e.g. https://mathoverflow.net/questions/18787/montagues-reflection-principle-and-compactness-theorem). The problem is, the proof that ZFC can do this takes more than ZFC, so ZFC can't prove "all my finite fragments are consistent." This is very weird, and takes some time to understand. 
