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This is false. Let $U=\mathbb N$ and let $S$ consist of all 2-element sets of the form $\{2n,2n+1\}$. Let $c$ be an ultrafilter extending the Frechét filter (so all finite sets are $\not\in c$). The statement $\exists D\in S\,\exists C\in c (C\subseteq D)$ is false, because the elements of $S$ are finite. And the statement $\exists C\in c, \forall D\in S (\text{card}(C\cap D)\le 1)$ is false because we can take $C=\mathbb N$, and there is an element of $D$ having more than one element.