In effect, you are asking if $\bigvee\limits_{i \in \omega}B_i = \bigcup\limits_{i \in \omega}B_i$, where the left hand side is the closure of the union, which is the join/supremum of the family $\{B_i\}_{i \in \omega}$ in the Boolean algebra of clopen sets of $X$, which we will write as $\mathrm{Clopen}(X)$. It is possible to characterize exactly when this happens.

We first consider the case when the family is not "genuinely infinite", in that there exists a finite set $I \subset \omega$ such that $\bigcup\limits_{i \in \omega}B_i = \bigcup\limits_{i \in I} B_i$. It follows that $\bigcup\limits_{i \in I} B_i$ is clopen, so the boundary of $\bigcup\limits_{i \in \omega}B_i$ is empty.

In the other case, suppose the family **is** "genuinely infinite", which is to say that for any finite set $I \subset \omega$, $\bigcup_\limits{i \in I}B_i$ is a proper subset of $\bigcup\limits_{i \in \omega}B_i$. We will show that the boundary of $\bigcup\limits_{i \in \omega}B_i$ is not empty.

We do this using Stone duality and the Boolean prime ideal theorem. Let's define $B = \mathrm{cl}\left(\bigcup_{i \in \omega}B_i\right) = \bigvee_{i \in \omega}B_i$. We are looking for a point $x \in X$ such that $x \in B$ but $x \not\in \bigcup_{i \in \omega}B_i$. By Stone duality, this is equivalent to finding an ultrafilter $u$ on $\mathrm{Clopen}(X)$ such that $B \in u$ but $B_i \not\in u$ for all $i \in I$. Let $\mathcal{F}$ be the principal filter on $B$ in $\mathrm{Clopen}(X)$, *i.e.* the set of clopen sets containing $B$, and let $\mathcal{I}$ be the ideal generated by the set $\{B_i\}_{i \in \omega}$ in $\mathrm{Clopen}(X)$, *i.e.* the set of clopen sets $G$ such that there exists a finite set $I \subset \omega$ such that $G \subseteq \bigcup_{i \in I}G_i$. Our assumption that the join is "genuinely infinite" ensures that $\mathcal{F} \cap \mathcal{I} = \emptyset$. So by the Boolean prime ideal theorem there exists an ultrafilter $u$ such that $\mathcal{F} \subseteq u$ and $u \cap \mathcal{I} = \emptyset$. Therefore $B \in u$ and $B_i \not\in u$ for all $i \in \omega$, as required.

i.e.Stone spaces) are compact Hausdorff spaces). $\endgroup$ – Robert Furber Sep 13 '18 at 1:52i.e.the Boolean algebra is countable). An infinite $\sigma$-complete Boolean algebra contains a subalgebra isomorphic to $\mathcal{P}(\omega)$ so is never countable. Therefore a Stone space that is a $\sigma$-space is second countable iff it is finite. (Also, write @ before the user name if you want to reply to a comment -- then the user you are replying to will be notified of it). $\endgroup$ – Robert Furber Sep 14 '18 at 10:03