$\displaystyle Let \hspace{1 mm} S = \{x:x^Tx \hspace{1 mm} $\leq$1\}.\hspace{1 mm}Prove\hspace{1 mm}that\hspace{1 mm}the\hspace{1 mm}extreme\hspace{1 mm}points\hspace{1 mm}of\hspace{1 mm}S\hspace{1 mm}are\hspace{1 mm}the\hspace{1 mm}points\hspace{1 mm}on\hspace{1 mm}its\hspace{1 mm}boundary.$

So $\displaystyle x$ is an extreme point of $\displaystyle S$ iff it is a basic feasible solution.

Would proof by contradiction be appropriate? Assume that the extreme points of $\displaystyle S$ are not on the boundary?