
A set is open iff for every point , there exists a such that the neighborhood surrounding is completely contained within (is a subset of) .

A point is a limit point of a set iff given any , all neighborhoods intersect the set in some point other than . Note that need not be a point in .
Let be a point in . Since is open, there is a number such that .
We now show is a limit point of . To that end, let be given. We may assume, without loss of generality, that . Now, , so . Since intersects in a point other than , we conclude is a limit point.
If , then , so . Openness already gives you an open interval around , so you only need to worry about the case where is smaller than that.