Is a broken (Euclidean) ball still convex?

To simplify notation, I'll assume, without loss of generality, that the center of the ball is at the origin. Suppose, toward a contradiction, that you have a counterexample in some high dimension. Say your set contains $p$ and $q$ but not some point $tp+(1-t)q$ on the line segment joining $p$ to $q$. Then this is also a counterexample in the $2$-dimensional subspace spanned by (the vectors from the origin to) $p$ and $q$. So, once you know the result for $2$ dimensions, it follows for all higher dimensions.