Another way to see that the difficulty there is the each appearance of "provable" in quotes refers to the formal Pvbl relation, which only resembles the real provability relation in certain respects.
In particular, there are consistent theories $T$ that extend PA such that $T \vdash \text{Pvbl}_T(\phi)$ for every formula $\phi$. But of course $T$ does not actually prove every formula $\phi$, because $T$ is consistent.
In the argument in the question, there are several steps that mix real provability with formalized provability. Let's assume that by "false" statement you mean a statement disprovable in the theory.
For example, step 2 begins with the fact that, if $P$ is false, then $P$ is not provable, so by contraposition, if $P$ is provable then $P$ is true. But this does not mean the theory can prove "if $P$ is provable then $P$ is true", which is the formula $\text{Pvbl}(P) \to P$. We cannot expect the theory to prove $\lnot P \to \lnot \text{Pvbl}(P)$, either, because as above there are consistent theories in which $\text{Pvbl}(P)$ is provable for all $P$.
The way to avoid this sort of thing is to be more explicit about writing Pvbl for the formalized provability relation, rather than using "provable" in quotes, which tends to cause confusion.
In particular, Löb's theorem tells us that if $T \vdash \text{Pvbl}(P) \to P$ then $T \vdash P$. That is stronger than the claim that if
$T \vdash \text{Pvbl}(P) \to P$ then $T \vdash \text{Pvbl}(P)$. The distinction between these is not nearly as clear when "provable" is used for both "$\vdash$" and "Pvbl".