It is my understanding that for a theory not to prove the negation of its Gödel sentence $G$ the theory has to be $\omega$-consistent. I have seen the definition of $\omega$-consistency, though I still can't quite get why it is required. For instance, the informal argument for the undecidability of $G$ goes like: "Assume $T$ is consistent. Suppose $T$ proves $G$. Then let $k$ be the Gödel number of this proof. Then $T \vdash Pk\sharp G$ and so $T$ proves $\exists x Px\sharp G = \neg G$ and $T$ is inconsistent. Now suppose $T \vdash \neg G$. * Then for some $k$, $k$ is the Gödel number of a proof of $G$, thus $T \vdash G$ and $T$ is inconsistent."
Now, I imagine the mistake is in the part marked by the asterisk, as in an $\omega$-inconsistent theory we can prove $\exists x \phi(x)$ without ever proving $\phi(k)$ for some $k$. But I still can't seem to make it click. Foregoing that part of the argument we just have that it is not necessary that the theory not prove $\neg G$, but how can we be sure that it does prove? Particularly, how does $PA + \neg Con(PA)$ prove $\neg G$, for instance? I can see how it would prove e.g. "there is a proof of a contradiction from $PA$" despite for all $k$ proving "$k$ is not a proof of a contradiction from $PA$", as the former is an axiom, but I can't picture what a deduction of $\neg G$ would be like. If anyone can show me, I think that would help. Or even help me in a different way if showing one explicitly is not possible (nonstandard length?).
I have searched and found another question here on SE on the same thing, by the way. Though the answers there didn't do it for me, unfortunately. I wasn't sure of the etiquette - e.g. should I revive that question? (Some communities frown upon this) - so I decided to post a new one.
By the Godel sentence, I assume you mean the $G$ which basically says, "I am not provable in $PA$."
Let $T=PA+\neg Con(PA)$. Remember the following basic fact of first-order logic: an inconsistent theory proves everything.
The crucial fact is that the fact above is provable in $PA$! Specifically, $PA$ proves "If I am inconsistent, then I prove $\varphi$" for every sentence $\varphi$. This is because $PA$ is strong enough to reason about basic manipulations of proofs, so e.g. $PA$ is strong enough to reason as follows:
(If you really want to get nitty-gritty, you can actually cook up a computable function $f$ such that $PA$ proves that if $k$ is the Godel number of a sentence, and $m$ is the Godel number of a $PA$-proof of $0=1$, then $f(k, m)$ is the Godel number of a proof of the sentence with Godel number $k$.)
Now, $G$ happens to be a sentence in the language of $PA$; so - according to $PA$ - if $PA$ is inconsistent, then $PA$ proves $G$.
Well, $T$ proves that $PA$ is inconsistent, and $T$ proves that if $PA$ is inconsistent then $PA$ proves $G$ (since $T$ contains $PA$), so $T$ proves "$PA$ proves $G$." But this is exactly $\neg G$!