I'm attempting this problem from Kunen:
I'm trying to do it by a direct combinatorial argument. Namely, let $C$ be the set of countable ordinals for which there exist such an $\omega$-chain. I want to see that $C$ is club. The fact that $C$ is unbounded is easy. I'm stuck on checking that it's closed. It's enough to consider the case $D_0\supset D_1\supset\dots$.
If $\gamma$ is a limit point of $C$, we can pick an increasing sequence $\langle\gamma_k:n<\omega\rangle$ which converges to $\gamma$, with each $\gamma_k\in C$. So we can choose $\langle p_n^k:k<\omega\rangle$ witnessing $\gamma_k\in C$. I tried to extract some sequence by a diagonal process, but the problem is that, in passing from $p_n^k$ to $p_m^{k+1}$, the later need not end-extend the former, which is part of the requirement. And we can't carelessly apply the density of the $D_n$ because we might add sets whose maximum is above $\gamma$.
I suspect you are misinterpreting the hint, it says you need to produce a particular club with the required property, not that the club to consider is defined by that property.
In particular it might be beneficial to instead show that the set $X$ of all $\delta\in \omega_1$ that are limit points of $S$ and for which there is a countable $A\subset \mathbb{P}_S$ such that
$(A,\le)$ is atomless with $p \in A$,
For each $n\in \omega$, $A\cap D_n$ is relatively dense in $A$,
For every $q \in A$ and $\alpha \in \delta \cap S$, $q\cap \alpha \cup \{ \alpha \} \in A$, and
$\{\max(q): q\in A\} \subset \delta$.
contains a club.