Suppose $S\subset \omega_1$ is a stationary set of limit ordinals. For each $\alpha\in S$, let $\langle \beta_n^\alpha:n\in\omega\rangle$ be an increasing sequence cofinal in $\alpha$, i.e. $\beta_n^\alpha<\alpha$ and $\beta_n^\alpha\nearrow\alpha$.
Given $T\subset S$, I'd like to define an injective function $g:T\to\omega_1$ such that, for each $\alpha\in T$, $g(\alpha)\in \{\beta_n^\alpha:n\in\omega\}$.
For $T$ countable, such a $g$ exists by an easy recursion. For $T$ stationary, there can be no such $g$ by Fodor's Lemma. I'm stuck on the case when $T$ is non-stationary.
Some thoughts:
If $T$ is not countable and we're trying to define $g$ by recursion on $\alpha\in T$, there is in general no reason why $\{g(\gamma):\gamma <\alpha\wedge \gamma\in T\}$ won't be unbounded in $\alpha$.
If $T$ is non stationary, there is a club $C$ with $C\cap T=\emptyset$. For each $\alpha\in T$, there is some $n_\alpha\in\omega$ such that $\beta_\alpha^n\not\in C$ for every $n\ge n_\alpha$.
The following is also true (although not necessarily useful) there exists some fixed $n\in\omega$ such that, for every $\eta<\omega_1$, $\{\alpha\in S:\beta_n^\alpha\ge \eta\}$ is stationary. Considering the (regressive) function $\alpha\mapsto \beta_n^\alpha$, for each $\eta<\omega_1$ we have (again by Fodor) some $\delta_\eta\ge \eta$ with $\{\alpha\in S:\beta_n^\alpha=\delta_\eta\}$ is stationary.
The argument in the countable case actually shows something stronger:
This lets us approach the whole problem by chopping it into blocks. Suppose $T\subseteq\omega_1$ is nonstationary. Let $C\subseteq\omega_1$ be a club disjoint from $T$, and enumerate $C$ in order as $\langle\gamma_\eta\rangle_{\eta<\omega_1}$. For $\eta<\omega_1$ we let the $\eta$th block of $T$ be the set $$S_\eta= T\cap [\gamma_\eta,\gamma_{\eta+1}].$$ Now apply $(*)$.
(The key point is that when we chop $T$ into blocks, we don't lose anything in the "gaps." When $T$ is stationary, there is no way to partition $T$ in this manner.)