 Solution

Suppose $S+A$ proves every specified reflection instance. Use only the instance with $X=\neg A$. The deduction theorem givesClassical propositional reasoning turns this into $S⊢\square \neg A\to \neg A$: if $\neg A$ already holds the implication is true, while otherwise $A$ holds and the displayed theorem supplies it. Löb's theorem now yields $S⊢\neg A$, making $S+A$ inconsistent. Therefore no sentence consistent with $S$ can supply all the local reflection instances in a one-sentence extension. The proof uses just one well-chosen instance, and never applies necessitation to a derivation with the extra assumption still open.