set theory - Club sets and substructures

The idea in my previous answer can, I think, be upgraded to solve the whole problem, as follows. Again, fix Skolem functions for $mathcal A$ as given by the well-ordering $<$, and again let $D$ be a set of fewer than $kappa$ ordinals $delta$, each of which is $sup(kappacap M_delta)$ for some $M_deltaprecmathcal A$ with $|M_delta|<kappa$. I need to show that $sup(D)$ is also in $C_{mathcal A}$. For each $deltain D$, define $N_delta$ to be the Skolem hull of $kappacapbigcup_{xiin D, xileqdelta}M_xi$. With appropriate gratitude for the hypothesis that $kappa$ is regular, note that $N_delta$ is an elementary submodel of $mathcal A$ of size $<kappa$ and that the sequence $langle N_deltarangle$ is an elementary chain. Let $N$ be its union, and note that it, too, is an elementary submodel of $mathcal A$ of size $<kappa$. So it suffices to show that $sup(D)=sup(kappacap N)$. The $leq$ direction here is obvious, as $N$ includes $kappacap M_delta$ for each $deltain D$. To complete the proof, suppose the $geq$ direction failed. Then we would have $sup(D)<sup(kappacap N)$, so there would be an ordinal $alphainkappacap N$ with $sup(D)leqalpha$. By construction, we would have some $deltain D$ with $alphain N_delta$, and so $alpha$ would be of the form $f(vecbeta)$ for some Skolem function $f$ and some ordinals $beta_i$ in $kappacap M_{xi_i}$ for certain $xi_ileqdelta$. For each $i$, we have $beta_i<xi_ileqdelta$, and, since there are only finitely many $i$ (as Skolem functions are finitary), we can find $gamma<delta$ with all $beta_i<gamma$. Increasing $gamma$ if necessary, we can arrange that $gammain M_delta$. In $mathcal A$, we can define the function $g$ sending each ordinal $nu<kappa$ to the supremum of all $f(veceta)<kappa$ for $eta$ bounded by $nu$; the values of this function are $<kappa$ by regularity. As an elementary submodel of $mathcal A$, $M_delta$ is closed under $g$ and, in particular, contains $g(gamma)$. But (again by elementarity) $g(gamma)$ majorizes $f(vecbeta)=alpha>sup(D)geqdelta$. That contradicts the fact that $delta$ is the supremum of $kappacap M_delta$, and this contradiction completes the proof.