Interior of a convex set is convex [duplicate]
Let $U= S^\circ$. Fix $0<t<1$. We have $tS + (1-t)S \subseteq S$ by convexity, so $tU + (1-t)U \subseteq S$. But $tU$ is open, so $tU+(1-t)U$ is open (exercise, sum of an open set and any set is open), and thus $tU + (1-t)U \subseteq S^\circ = U$, and hence $U$ is convex.