How to prove the Gödel sentence is true (in the metalanguage), assuming only consistency of a theory.

Question

Let $T$ be a consistent, axiomatizable extension of $Q$ (Robinson or Minimal arithemthic shouldn't matter).
We construct the Gödel sentence by using the diagonal lemma as: $\vdash_T G_T \Leftrightarrow \neg \exists y Prf_T(\ulcorner G_T \urcorner,y),$ where we can take $Prf_T$ to be rudimentary.
In the usual way you can assuming $\omega$-consistency prove that $G_T$ is undecidable in $T$.
What I am wondering is how would you go about arguing that the Godel sentence is true, assuming only consistency (or $\omega$-consistency I guess) without assuming the theory $T$ is sound? I know this argument cannot be done inside of $T$, I am looking for a metalanguage argument for why it is true.
The way I've seen this argued is: the Gödel sentence is provably equivalent to $\forall y \neg Prf(\ulcorner G_T \urcorner ,y)$ such sentences are provably false whenever they are actually false (in the standard interpretation). If false, there is a number $n$ such that $\vdash_T Prf_T(\ulcorner G_T \urcorner, \mathbf{n})$, but this contradicts the first Incompleteness theorem.
I might be wrong but I feel like this argument uses soundess of the theory.
I don't understand why it wouldn't be possible that $\vdash_T \neg G_T \Leftrightarrow \exists y Prf_T(\ulcorner G_T \urcorner,y)$ without this equivalence being true, I guess it somehow contradicts consistency but I don't see how.