"Theorem": There is no $\omega$th inaccessible cardinal.
"Proof": Assume ZFC. Let $\kappa_n$ be the $n$-th inaccessible cardinal; since $V_\kappa$ is a model of ZFC for inaccessible $\kappa$, $V_{\kappa_n}$ is a model of ZFC + "there are $n-1$ inaccessible cardinals."
Now suppose there are $\aleph_0$ many inaccessible cardinals. Then in particular we can take the $\omega$-th inaccessible $\kappa_\omega$, and there will still be $\aleph_0$ many inaccessibles below it. Thus $V_{\kappa_\omega}$ is a model of ZFC+"there are $\aleph_0$ many inaccessible cardinals", and so by the (second) incompleteness theorem that theory is inconsistent. $\square$
But this "result" is literally too good to be true. So where am I wrong here?
Edit: Yep, I realize now what the fallacy is. Daniel Fischer and GME are correct - the existence of $\aleph_0$ many inaccessibles does not imply the existence of an $\omega$th inaccessible, just as the fact that there are $\aleph_0$ many natural numbers does not imply the existence of the ordinal $\omega$.
In other words, there can be models of ZFC + "there are $\aleph_0$ many inaccessibles" such that every inaccessible has finitely many inaccessibles below it, so the argument fails. In fact $V_{\kappa_\omega}$ is just such a model: it knows of $\kappa_n$ for all finite $n$, but it sees $\kappa_\omega$ as a proper class.