So, we're in a metric space, right? Assume first that is open but that this does not hold. Then, there exists and a sequence failing to eventually in . Clearly then any neighborhood is going to contain all but finitely many of the and so, by assumption, an element of . Thus, there does not exist a neighborhood of contained in , which contradicts openess. Conversely, if is open and you know that there exists a neighborhood of with . But, since we have by definition that all but finitely many .