We will show that if "U" ⊆ "X" is non-empty and open, and if "x" is a point of "X", then there is a neighbourhood "V" ⊂ "U" whose closure does not contain "x" ("x" may or may not be in "U"). Choose "y" in "U" different from "x" (if "x" is in "U", then there must exist such a "y" for otherwise "U" would be an open one point set; if "x" is not in "U", this is possible since "U" is non-empty). Then by the Hausdorff condition, choose disjoint neighbourhoods "W" and "K" of "x" and "y" respectively. Then "K" ∩ "U" will be a neighbourhood of "y" contained in "U" whose closure doesn't contain "x" as desired.