Going out on a limb, I assume you have a topological (or metric, or normed) space X and the set in question is of the form S={x in X | x has property P}. And that P is a local property in the sense that if x has the property P, then so does every point in a suitable open neighborhood of x. Then this neighborhood would, by definition, be contained in S. And since this is true of every point of S, there is an open neighborhood around every point of S, which is also contained in S. Which makes S open.