This is an exercise from Munkres' Topology. I have proved there exist open sets $U$ and $V$ of $X$ containing $C$ and $D$, respectively. Now I need to show $$\bigcap_{A\in \mathcal{A}} A- (C \cup V)$$ is non-empty. I think I need to use nested set property to prove this. Because I want to have $[A_1- (C \cup V)] \subset [A_2- (C \cup V)] \subset \cdots$, can I label $A\in \mathcal{A}$ by $A_i$ for $i\in \mathbb{Z}_+$?
Nested set property:
First things first, show $A \setminus (U \cup V)$ is non-empty for all $A \in \mathcal{A}$. If $A \setminus(U \cup V) = \emptyset$ then $A \subset U$ or $V$ as $A$ is connected and thus $Y \subset U$ or $V$ and $C, D$ would no longer be a partition. This implies that $A \setminus (U \cup V)$ is a non-empty compact set for each $A \in \mathcal{A}$.
Lemma: If $I$ is simply ordered and $(C_i)_{i \in I}$ is a nested with respect to the order of $I$ then $\cap_{i \in I} C_i \neq \emptyset$.
Proof: Suppose $\cap_{i \in I} C_i = \emptyset$ and choose $j \in I$. Define for $i \geq j$ the open subset $U_i = C_j \setminus C_i$ of $C_j$, then $$ U:= \cup_{i \geq j} U_i = C_j \setminus \cap_{i \geq j} C_i = C_j \setminus \cap_{i \in I} C_i = C_j.$$ As $(U_i)_{i \geq j}$ is a cover of the compact set $C_j$ then there exists $U_{i_1} \supset \ldots \supset U_{i_n}$ such that $$C_j = \cup_{k=1}^n U_{i_k} =C_j \setminus \cap_{k=1}^n C_{i_k} = C_j \setminus C_{i_n}.$$ Since $C_{i_n} \subset C_j$ this implies $C_{i_n} = \emptyset$ which contradicts our original assumption. QED.
We now apply the lemma to say $$(C \cup D) \setminus (U \cup V) =Y \setminus (U \cup V) = \cap_{A \in \mathcal{A}} A\setminus (U \cup V) \neq \emptyset$$which contradicts that $C \subset U$ and $D \subset V$.