The typical topologist's sine curve is defined as,
\begin{align}\label{1}\tag{1}& S = X \times Y = \{(0,y)| -1 \leq y \leq 1 \} \times \{(x, \sin (1/x))| x>0\}\\& X = \{(0,y)| -1 \leq y \leq 1 \}, \quad Y = \{(x, \sin (1/x))| x>0\}\end{align}
I want to modify this a bit by changing the range of $X$ to $-1 < y < 1$ so that,
\begin{align}\label{2}\tag{2}& S = X \times Y = \{(0,y)| -1 < y < 1 \} \times \{(x, \sin (1/x))| x>0\}\\& X = \{(0,y)| -1 < y < 1 \}, \quad Y = \{(x, \sin (1/x))| x>0\}\end{align}
Intuitively, we might seem to think that $X$ and $Y$ are disconnected due to the fact that they are disjoint $X \cap Y = \emptyset$. However, the definition for being connected is that, $X$ and $Y$ are connected only if we cannot write it as $X \cup Y = S$ where both $X$, $Y$ are nonempty open sets, and that $X \cap Y = \emptyset$. So, to argue that $X$ and $Y$ are disconnected due to $X \cap Y = \emptyset$ only is not enough. We also have to show that there is a way to write $X \cup Y = S$ where both $X$, $Y$ are nonempty open sets.
Now, $X$ in $\eqref{1}$ is closed since it contains all of its boundary points, while $X$ in $\eqref{2}$ is neither closed nor open. First question, is it correct to say that at least for motivation but not proof, that $X$ and $Y$ certainly cannot be disconnected since $X$ in both cases is not open? I have added $\eqref{2}$ to emphasize what I'm thinking about the arguments involving the topologist's sine curve involves $X$ not being open such that whatever proof works for $\eqref{1}$ should also work for $\eqref{2}$. In other words, whatever proof for the topologist's sine curve, it doesn't matter whether $-1 \leq y \leq 1$ or $-1 < y < 1$.
There is a proof for the connectedness of $S$ in $\eqref{1}$ which goes as follows (I'm not going to write the whole proof since that's not where my question lies): We assume (and know) that $Y$ is connected, we want to show that the closure of $Y$ labeled $\bar{Y}$ is connected. The closure $\bar{Y}$ will of course include $X$ so that $S = \bar{Y}$. Let us suppose that $\bar{Y}$ is disconnected, so there should be $A$, $B$ both nonempty open sets such that $A \cup B = \bar{Y}$ and $A \cap B = \emptyset$. The proof can be done to show in the end that either $A$ or $B$ must be empty, but this contradicts the assumption that both $A$, $B$ must be nonempty so $S = \bar{Y}$ must be connected.
Second question, is it necessary that $A = X$ and $B = Y$? If yes, then wouldn't the assumption that $A$, $B$ both nonempty open sets be at odds with the already given fact that $X$ is not open in both cases $\eqref{1}$ and $\eqref{2}$?