Does closure against countable union survive union of classes?
The answer to (3) is yes. Suppose that $S \subseteq \mathcal{C} \cup \mathcal{C}'$. Then $S$ is countable, because $\mathcal{C}\cup\mathcal{C'}$ contains only decidable languages of which there are only countable many. Now let $T = S \cap \mathcal{C}$, $T' = S \cap \mathcal{C}'$. By assumption $L = \bigcup T \in \mathcal{C}$, $L' = \bigcup T' \in \mathcal{C}'$. Since $L,L'$ are decidable, $L \cup L'$ is decidable.