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 $(a - A)^2 + (b - B)^2 + (c - C)^2 \leq (a^2 + b^2 + c^2) + (A^2 + B^2 + C^2) = 2r^2$, so the distance is at most $r\sqrt{2}$, as required.