Collection of all open sets whose closure is contained in a cover is a cover (of a regular space) Let $X$ be a regular space. Let $\mathcal{A}$ be an open cover of $X$. We define $\mathcal{B}$ as the collection of all open sets $U$ such that $\overline{U}$ is contained in an element of $\mathcal{A}$. How to prove that $\mathcal{B}$ covers $X$?
(My attempts to bring in regularity failed so far)

For $x\in X$ there is some $V\in\mathcal A$ with $x\in V$ or equivalently $x\notin V^c$. 
Set $V^c$ is closed so the regularity of $X$ ensures the existence of open sets $U,W$ s.t. $x\in U$ and $V^c\subseteq W$ and $U\cap W=\varnothing$. 
Set $W^c$ is closed, so $U\subseteq W^c$ implies that $\overline{U}\subseteq W^c\subseteq V\in\mathcal A$. 
This shows that $U\in\mathcal B$ so we have: $x\in U\in\mathcal B$.
Proved is now that $\mathcal B$ covers $X$.
