A point y is on the boundary of A iff for any for any open set U containing y we have that U intersects both A and X-A. If y is on the closure of A then y is in A or the boundary of A. If y is in A then every open set containing y obviously intersects A. And if y is on the boundary of A then any open set containing y intersects A (by what was said above). Thus, any open set containing y intersects A. Try doing the backwards direction.