Why A→B∉Γ implies Γ∪{A}$\not\vdash$B

Question

The setting is intuitionistic propositional logic. Γ is a consistent disjunctive(A∨B∈Γ⇒A∈Γ or B∈Γ) theory. I can't figure out why the claim stated in the title is correct. Let Γ be the smallest consistent disjunctive theory which contains p→(p→q). Then p∉Γ,p→q∉Γ but Γ∪{p}⊢p→q so Γ∪{p}⊢q. If the claim in title is right then my reasoning must be wrong but I can't see why.

The picture below is the original problem, if it helps.

Answer 1

If $\Gamma \cup \{ A \} \vdash B$, then by $\to$-intro :

$\Gamma \vdash A \to B$.

But $\Gamma$ is regular, and thus closed, and so : $A \to B \in \Gamma$ : contradiction.