Definition:
A space $X$ is a Urysohn space iff whenever $x \neq y$ in $X$ there are nhoods of $U$ of $x$ and $V$ of $y$ such that $\overline{U} \cap \overline{V} = \emptyset$.
I want to show that every regular, T_1 space is Urysohn.
My attempt:
Let $x,y \in X$ be distinct points. Now consider the open set $X \setminus \{y\}$. By regularity we can find an open set $U$ such that $x \in U \subseteq \overline{U} \subseteq X \setminus \{y\}$. Now similarly by considering the open set $X \setminus \{x\}$ and using regularity again we can find an open set $V$ such that $y \in V \subseteq \overline{V} \subseteq X \setminus \{x\}$.
Now from where I don't think we can conclude $U$ and $V$ have disjoint closures. Can we? I'm stuck here, can you please help?