Understanding a detail of a proof about topological compactifications

Question

I don't completely the following proof from Van Douwen's paper. My exact questions are below the proof.

Van Douwen proves that

If $X$ is a non-compact strongly zero-dimensional space in which every nonempty clopen subspace is homeomorphic to $X$, then the only H-compactification (or "topological compactification") of $X$ is the Stone-Čech compactification.

The Proof:

Let $\gamma X$ be an arbitrary H-compactification of $X$. In order to show that $\gamma X$ is the same as $\beta X$ we prove that every clopen subset of $X$ has an open closure in $\gamma X$, which will, according to one of the equivalent characteristics of the Stone-Čech compactification, lead to the desired conclusion.

Denote again by $\overline{\:}$ the closure operator in $\gamma X$. If $U$ is a clopen subset of $X$, we can assume $\emptyset \neq U \neq X$. Then we can find a nonempty clopen subset $V$ of $X$ such that $\overline{U} \cap \overline{V} = \emptyset$. $U$ and $V$ are indeed homeomorphic. Then there exists an automorphism $h$ of $X$ such that $h(U) = V$ and $h$ sends any $x \not\in U \cup V$ to itself.

By the assumption, $\gamma X$ is a H-compactification, so we can define an extension of $h$ over $\gamma X$. Denote such extension by $\gamma h$. Since the intersection of closures of $U$ and $V$ is assumed to be empty and $h$ maps every $x \in U$ to $y \in V$, we see that $\gamma h$ satisfies

$\overline{U} \cap (\gamma h)(\overline{U}) = \overline{U} \cap \overline{V} = \emptyset$.

Observe that from our definition of $h$, $\gamma h(x) = x$ for each $x \in \overline{(X \setminus (U \cup V))}$. Therefore we have

$\overline{U} \cap \overline{(X \setminus U)} = (\overline{U} \cap (\gamma h)(\overline{U})) \cup (\overline{U} \cap \overline{(X \setminus (U \cup V))}) = \emptyset$.

Hence the closure $\overline{U}$ is open in $\gamma X$.

My questions are:

1. I don't understand the last sentence in bold. How do we obtain the conclusion that the $\overline{U}$ is open in $\gamma X$? and Did we somehow use the properties of $X$ stated in the claim? Is it because since $U$ is homeomorphic to the whole $X$? (I still don't see how this would imply the last sentence though.)

2. How is the strong zero-dimensionality used in the proof?

Definition of H-compactification (or "topological compactification") is that every automorphism on the original space can be continuously extended over the compactification.

Answer 1

1. Because you now have that $\overline{U}$ is the complement of $\overline{X\setminus U}$ in $\gamma X$ (which is a closed set) the set $\overline{U}$ is open.
2. In the very first paragraph of the proof: strong zero-dimensionality means that disjoint zero-sets are separated by clopen sets, and the fact that clopen sets in $X$ have clopen closures in $\gamma X$ implies that disjoint zero-sets in $X$ have disjoint closures in $\gamma X$ (see Corollary 3.6.2 in Engelking's book).