Rudin Theorem 2.47 - Connected Sets in $\mathbb{R}$

A response to iii) of my question with the help of Alex Becker's answer. I'd appreciate if anyone could check my understanding of the rest of the proof.

In particular, $x \leq z < y.$

We have $z \in \overline{A}, z \in [x,y]$ and $x \in \overline{A}$ so $x \leq z$. As $z \not\in B$ and $z \in [x,y]$ we must have $z < y$.

If $z \not \in A$, it follows that $x < z < y$, and $z \not\in E$.

As $z \in [x,y]$ and $z \neq x$ (since $z \not\in A$) we have $x < z < y$ and $z \not\in A \cup B = E$.

If $z \in A$, 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 $z_1 \not \in E$.

Since $z \not\in \overline{B}, z \in [x,y]$ and $y \in \overline{B}$ we have $z < y$. As $z \not\in \overline{B}$, $z$ must be in the complement of $\overline{B}$ which is an open set in $\mathbb{R}$. Therefore, there exists $r > 0$ such that $(z - r, z + r) \subset \mathbb{R} - \overline{B}$. So taking $z_1 \in (z,z + r)$ we have $z < z_1 < y \text{ and } z_1 \not\in \overline{B} \supset B$ and clearly, $z_1 \not\in A$ as it would contradict the definition of $z$. Therefore, $x < z_1 < y$ and $z_1 \not\in A \cup B = E$.

i) $z$ is the supremum of $A$, restricted to the region $[x,y]$. This is necessary because we need $z\in [x,y]$.

ii) $\overline{A\cap [x,y]}\subseteq \overline{A}$, so if $z\in \overline{A\cap [x,y]}$ then...