 Solution

Assume $S⊢B_{S}X\to X$ and put $T=S+\{\neg X\}$. Then $T⊢\neg B_{S}X$. The given extension-provability implication, with $Y=\perp$, givesThe provable classical implication $(\neg X\to \perp )\to X$, together with distribution, gives $S⊢B_S(\neg X\to \perp )\to B_{S}X$. HenceIf $T$ were consistent, this would violate its Gödel second incompleteness theorem. Thus $T$ is inconsistent. The deduction theorem for propositional logic and double-negation elimination give $S⊢\neg \neg X$, hence $S⊢X$. Therefore the second incompleteness theorem for $S+\{\neg X\}$ yields Löb's theorem for arbitrary $X$.