My attempt: Suppose $C\cup D$ is a separation of $Y\cup A$. Then, since $Y$ is connected, we can say, without loss of generality, $Y\subset D$. Then, we can write $X=C\cup (D\cup B)$.
This is where I get stuck. I want to show $C\cup(D\cup B)$ is a separation of $X$, and get a contradiction, but I'm not sure exactly how I would go about showing $C$ is open in $X$ and $D\cup B$ is open in $X$.
They clearly form a partition of $X$, but how can we be certain they are open?
All we know is $C,D$ are open in the subspace topology of $Y\cup A$ and $B$ is open in the subspace topology of $X-Y$.
Once this is resolved, the case for $Y\cup B$ is identical.
It's a slog, but a picture will help. I drew a picture first then followed my nose. Picking up on your idea, let $X − Y = A\cup B$ be a separation of $X − Y$ and $Y\cup A = C\cup D$ be a separation of $Y\cup A$.
We will use the fact that if $A$ and $B$ separate a subpace $Z$ of $X$, then $\overline A\cap B=A\cap \overline B=\emptyset.$
Then, we have
$\overline A \subseteq X − B;\ \overline B\subseteq X − A;\ \overline C\subseteq X − D;\ \overline D\subseteq X − C;\ $and of course, $Y \cup A \cup B = B \cup C \cup D=X.$ We use these facts freely in what follows.
$Y$ is connected so wlog $Y\subseteq C$. Then, it's easy to see that $D\subseteq A$, so $\overline D\subseteq X-B.$
Furthermore, $\overline {B\cup C}=\overline B \cup \overline C \subseteq (X − A) \cup (X − D) = (B \cup Y ) \cup (B \cup C) \subseteq B \cup C.$ Similarly, you can show that $\overline D\subseteq D$.
Thus, $B\cup C$ and $D$ are closed subspaces, and since $X=B\cup C \cup D$, they form a separation on $X$ and we have our contradiction.