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I have the following proof, but I don't understand one of the steps:

Theorem 4.4. A Boolean algebra is complete iff its Stone space is exlremally disconnected.

Proof. Identify the given Boolean algebra $B$ with the clopen algebra of $\mathsf{S}B$. Suppose that $B$ is complete, and let $U$ be an open set in $\mathsf{S}B$. Let $\mathscr{A}$ be the family of all members of $B$ included in $U$; then, since $B$ is a base for $\mathsf{S}B$, we have $U = \bigcup \mathscr{A}$. Since $B$ is complete, $\mathscr{A}$ has a supremum $V$ in $B$ which must by definition be a clopen subset of $\mathsf{S}B$. We claim that $\overline{U} = V$. Since $V$ is an upper bound for $\mathscr{A}$, certainly $U = \bigcup \mathscr{A} \subseteq V$, so that $\overline{U} \subseteq V$ since $V$ is closed. If $V - \overline{U} \neq \emptyset$, then $V—\overline{U}$ is a non-empty open set which must, since $\mathsf{S}B$ is a Boolean space, include a non-empty clopen set $W$. But then $V-W$ is a clopen set which includes $\bigcup \mathscr{A}$ and is properly included in $V$. This contradicts the choice of $V$ as the supremum of $\mathscr{A}$. Therefore $\overline{U}=V$ as claimed, so a fortiori $V$ is open.

Conversely, suppose that $\mathsf{S}B$ is extremaily disconnected, and let $\mathscr{A}$ be a subfamily of $B$. Then $U = \bigcup\mathscr{A}$ is an open subset of $\mathsf{S}B$ (recall that we are identifying $B$ with $\mathsf{CS}B$!) and so, since $\mathsf{S}B$ is extremally disconnected, $\overline{U}$ is clopen and hence in $B$. We claim that $\overline{U} = \bigvee \mathscr{A}$ in $B$. Certainly $\overline{U}$ is an upper bound for $\mathscr{A}$ in $B$; on the other hand, if $V$ is a member of $B$ which includes each member of $\mathscr{A}$, then $U = \bigcup \mathscr{A} \subseteq V$ so that $\overline{U} \subseteq V$ since $F$ is clopen. Therefore $\overline{U} = \bigvee \mathscr{A}$ as claimed, and $B$ is complete. $\blacksquare$

I do not understand the argument that why $V - \overline{U}$ is open set. and why should it include $W$ clopen set.

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Since $V$ is clopen and the closure of $U$, $\overline U$, is closed, then as $V-\overline U=V\cap(SB-\overline U)$ by definition of set difference, then $V-\overline U$ is the finite intersection of two open sets, which is open by definition. In a topological space with an clopen base, any open set is the union of clopen sets, and hence a non-empty open set will contain a non-empty clopen set, W say, as a subset.

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