II OBM

Suppose the result is false so that we can find 5 points with the distance between any two greater than $r\sqrt{2}$. Then the angle subtended by any two at the center of the sphere is greater than $90^{\circ }$. Take one of the points to be at the north pole. Then the other four must all be south of the equator. Two must have longitude differing by at most $90^{\circ }$.

It is now fairly obvious that these two points subtend an angle at most $90^{\circ }$ at the center. To prove it we may take rectangular coordinates with origin at the center of the sphere so that both points have all coordinates non-negative. Suppose one is $(a,b,c)$ and the other $(A,B,C)$. Then since both lie on the sphere $a^2+b^2+c^2=A^2+B^2+C^2=r^2$, and the square of the distance between them is $(a-A{)}^2+(b-B{)}^2+(c-C{)}^2\le (a^2+b^2+c^2)+(A^2+B^2+C^2)=2r^2$, so the distance is at most $r\sqrt{2}$, as required.