= Solution
The <club filter completeness> argument works for every regular uncountable $\kappa$. Let $\mu<\kappa$ and let $C_i$ be a <club set> for each $i<\mu$. Their intersection is closed. To prove it unbounded, start above any prescribed $\beta<\kappa$ and choose an increasing sequence $(\delta_n)_{n<\omega}$ so that $\delta_{n+1}$ lies above a chosen point of every $C_i$ greater than $\delta_n$. $\kappa$ is a <regular cardinal>, so the supremum of these $\mu$ choices remains below $\kappa$. The resulting countable supremum $\delta=\sup_n\delta_n$ likewise remains below $\kappa$. For each $i$, the chosen $C_i$ points are cofinal in $\delta$, so closure gives $\delta\in C_i$. Thus the intersection is a <club set>. Intersecting fewer than $\kappa$ members of the <club filter> still contains such an intersection of clubs. In particular, $\boxed{\mathcal D_{\omega_2}\text{ is }\aleph_2\text{-complete}}$.
It is not an <ultrafilter>. For a regular infinite $\theta<\kappa$, the set
$$
S_\theta^\kappa=\{\delta<\kappa:\operatorname{cf}(\delta)=\theta\}
$$
is stationary. Given a <club set> $C$, build in $C$ a strictly increasing continuous sequence of length $\theta$ and take its supremum $\delta<\kappa$. Closure gives $\delta\in C$, and the <cofinality of an increasing ordinal supremum> gives $\operatorname{cf}(\delta)=\theta$. This proves the <stationarity of ordinals of prescribed cofinality>.
At $\kappa=\omega_2$, the disjoint sets $S_\omega^{\omega_2}$ and $S_{\omega_1}^{\omega_2}$ are both stationary. A club contained in either $S_\omega^{\omega_2}$ or its complement would miss one of these stationary sets. Thus neither $S_\omega^{\omega_2}$ nor its complement belongs to $\mathcal D_{\omega_2}$, and
$$
\boxed{\mathcal D_{\omega_2}\text{ is not an ultrafilter}}.
$$
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