= Solution
A <kappa-filtration> is an increasing continuous sequence with union $A$ and $|A_\alpha|<\kappa$ at every stage. Intersect $S$ with the <club set> of nonzero <limit ordinals>. For each remaining $\alpha$, continuity gives $A_\alpha=\bigcup_{\beta<\alpha}A_\beta$, so choose $r(\alpha)<\alpha$ with $f(\alpha)\in A_{r(\alpha)}$. This is a <regressive function>. <Fodor lemma> gives a stationary <subset> $S_1$ and a fixed $\beta$ such that all these values lie in $A_\beta$.
Since $|A_\beta|<\kappa$, partition $S_1$ into fewer than $\kappa$ fibers of $f$. If every fiber were nonstationary, choose a <club set> avoiding each one. Their intersection is <club set> by regularity, contradicting stationarity of $S_1$. Thus one fiber is stationary, and
$$
\boxed{\exists a\in A\ \exists S'\subseteq S\text{ stationary}:\ f\upharpoonright S'=a.}
$$
This proves the <filtration form of Fodor lemma>; continuity at limit stages, the small size of each stage, and regularity of $\kappa$ are all used.
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