Let's show that B is a base for a given topology iff it includes a base at p for every point p. Given the former, place p in an arbitrary open set W, which is the union of base sets, hence p is contained in a base open set inside W. Conversely, if every p has a base, cover W with base open sets containing the points of W, and contained in W, and W becomes the union of base open sets, as is required by a base. If p is in the intersection of base open sets, p is contained in a base open set that is contained in the intersection, and that takes care of the base criterion.