$2.47$ Rudin non-connectedness in $\mathbb R$ - how do we formally prove that $z_1 \notin A$ ? Consider the cases I drew in these schemes

Question

This is theorem $2.47$ from Principles of Mathematical Analysis by W. Rudin (it's aiming to prove that in $\mathbb R$ a part $X$ is disconnected iff it doesn't verify $\forall a,b \in X$ such as $a<b$ and $\forall x$ such that $a<x<b$ we have that $x\in X$):

Proof: To prove the converse suppose that $E$ is not connected. Then there are nonempty separated sets $A$ and $B$ such that $A \cup B = E$. Pick $x \in A, y \in B$ and assume (without loss of generality) $x < y$. Define $$z = \sup(A \cap [x,y])$$ By Theorem 2.28, $z \in \overline{A\cap [x,y]}\subset \overline{A}$; hence $z \not\in B$. In particular, $x \leq z < y$.
If $z \not \in A$, with $z \not\in B$, it follows that $x < z < y$, and $z \not\in E$.
If $z \in A$, with $z \not\in B$, then $z \not\in \overline{B}$, hence there exists $z_1$ such that $z < z_1 < y \text { and } z_1 \not\in B$. Then $x < z_1 < y$ and $z_1 \not \in E$. $_\Box$

My question concerns the very end of the proof: how do we know that $z_1\notin A$ ?

I understand that even if $A$ and $B$ are "alternating" (like in the image below), by taking $z=\sup(A\cap [x,y])$ we manage to be inside a gap "between" $A$ and $B$, as I have shown in my scheme. I also assume that "alternating" separated sets $A$ and $B$ are the worst case.

I also understand that the situation where $A\cap [x,y]$ is like I have shown in the below figure is impossible. Namely because $y\in B$ and $x\in A$ and $A\cap B= \emptyset$. However, I can't prove it (that it is impossible) formally.

I want to emphazise that if the situation where $A\cap [x,y]$ is like I have shown in my drawing were to happen, we could not conclude that $z_1>z$ is not in $A$.

Thanks

PS: sorry for representing $A$ as a bar in my second scheme, I now realize a line would have been more appropriated.