$U$ and $V$ are disjoint non-empty open sets in a Hausdorff space $X$. If  $U$ and $V$ are disjoint non-empty open sets in a Hausdorff-space $X$, then can we claim that the closure of $U$ and the open set V are also disjoint? I think this would be true if $X$ was a metric-space but cannot see why should it be true in a Hausdorff-space. If it is not true, then what additional properties on the bare Hausdorff-space would make the conclusion possible?
Thanks.

I believe the answer is affirmative in any topological space.
Suppose $X$ is a topological space and that $U,V\subseteq X$ are open, nonempty and disjoint. We want to show $\overline{U}\cap V = \emptyset$.
This follows immediately from the fact that for $x\in X$, $x\in \overline{U}$ iff for every open set $W\subseteq X$ with $x\in W$, $W\cap U \neq \emptyset$.
To see this, suppose there exists an $x\in \overline{U}\cap V$. Since $V$ is open, and since $x\in \overline{U}$, we must have that $U\cap V \neq \emptyset$ which is a contradiction.
