# Existence of Specific Regular Open Set in Topological Space

I'm trying to show that for any topological space $$X$$ and any $$O\subseteq X$$ with $$O$$ an open set, there is an regular open set $$W$$ such that $$O \subseteq W$$ and $$O$$ is dense in $$W$$.

I'm not quite sure what "$$O$$ is dense in $$W$$" means. Does this typically mean that $$O$$ is dense in the subspace topology induced by $$W$$ or for any nonempty open set contained in $$W$$, $$O$$ meets it?

I appreciate any clarification

Edit: I believe that I can just let $$W =$$ Int(Cl($$O$$)) as the regular open set, but I'm still trying to show the density part.

$$O$$ is dense in $$W$$ means that $$O$$ is dense in the subspace topology induced by $$W$$
To show it for your suggestion $$W= \operatorname{Int}(\operatorname{Cl}(O))$$ : suppose $$U$$ is a non-empty open subset of $$W$$.
Suppose that $$O \cap U=\emptyset$$, it follows that $$U \cap \operatorname{Cl}(O)=\emptyset$$ too ($$O \subseteq U^\complement$$, so $$\operatorname{Cl}(O) \subseteq U^\complement$$, as the right hand set is closed, so $$U \cap \operatorname{Cl}(O) = \emptyset$$), and so $$U \cap \operatorname{Int}(\operatorname{Cl}(O)) = \emptyset$$ as well, but this just says $$U \cap W= \emptyset$$, which cannot be as $$\emptyset \neq U \subseteq W$$ by assumption. So contradiction and thus $$U \cap W \neq \emptyset$$, QED.