Follow the idea of user04651 in this post to prove this statement.
Lemma. Let be a normed space, open and . If , then the line segment
Proof. For every , there is a such that . Let be such that .
Now for every (we have discussed the intuition of this radius in this post), we have for some ,
As this is true for every , we conclude .
Proposition. Let be a normed space and a convex set, then .
Proof. The direction is clear.
Let , for the sake of contradiction suppose , then there is an open neighborhood of such that . Since , there is an such that .
Now being an element in , , fix an , by the lemma above the segment
We can pick an that is close enough to , so that , but then , a contradiction.