= Solution
The <Fodor lemma>, also called <Fodor's theorem>, states: if $\kappa$ is a regular uncountable <cardinal>, $S\subseteq\kappa$ is stationary and $f:S\to\kappa$ is regressive, meaning $f(\alpha)<\alpha$ for every nonzero $\alpha\in S$, then some fibre $f^{-1}(\{\xi\})$ is stationary.
Delete zero from $S$, which preserves stationarity. Suppose no fibre is stationary. For each $\xi<\kappa$, choose a <club set> $C_\xi$ avoiding that fibre. The <diagonal intersection>
$$
C=\{\alpha<\kappa:(\forall\xi<\alpha)\ \alpha\in C_\xi\}
$$
is a club. To prove unboundedness, start above an arbitrary bound and choose a strictly increasing sequence $\alpha_n$ with $\alpha_{n+1}$ in every $C_\xi$ for $\xi\le\alpha_n$. The intersection of these fewer-than-$\kappa$ clubs is club by <club filter completeness>. Put $\alpha=\sup_n\alpha_n<\kappa$. For any $\xi<\alpha$, every sufficiently late $\alpha_n$ lies in $C_\xi$, so its closure gives $\alpha\in C_\xi$. Thus $\alpha\in C$. For closure of $C$, let $\alpha<\kappa$ be a limit point of $C$; for each $\xi<\alpha$, the members of $C$ above $\xi$ witnessing that limit lie in $C_\xi$, and closure gives $\alpha\in C_\xi$.
Stationarity supplies $\alpha\in S\cap C$. Since $f(\alpha)<\alpha$, we have $\alpha\in C_{f(\alpha)}$, but that club was chosen to avoid the fibre containing $\alpha$. This contradiction proves the theorem. For completeness, <club filter completeness> follows by a similar diagonal sequence: cycle through fewer than $\kappa$ clubs over $\omega$ rounds, each time choosing a larger member of the next club. Regularity keeps the supremum below $\kappa$, and every club is visited cofinally often in the construction, so its closure contains the supremum. Their intersection is also closed.
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