Suppose that the family $\mathrm{RO}(X)$ of regular open subsets of $(X,\mathscr{O})$ is a basis of $X$. Let the density of $\mathrm{RO}(X)$ (considered as Boolean algebra) be $\aleph\_0$.


Does $X$ have to be second-countable? If not, what if we add regularity of $X$ (both $T\_1$ and $T\_3$ separation axioms)? If answers to both questions are negative, what is the maximal cardinality (relative to $|X|$) of the set of points not covered by dense countable $D\subseteq\mathrm{RO}(X)$?



Does $\aleph_0$-density of regular open algebra entail existence of countable basis?
The answer is no, not necessarily. 


For a counterexample, consider the [Sorgenfrey line](https://en.wikipedia.org/wiki/Lower_limit_topology), which is the topology on $\mathbb{R}$ with basis consisting of the half-open intervals $[a,b)$. These are each clopen and hence regular open in that topology. Furthermore, every such nonempty open interval contains a nonempty half-open interval $[p,q)$ where $p,q\in\mathbb{Q}$, and so the regular open algebra is $\aleph\_0$-dense as a Boolean algebra. The Sorgenfrey line is normal but not second-countable. Thus, it serves as a counterexample to your question. 


For the last part of the question, this gives an example where the cardinality of the excess not covered has size continuum.


