Solution:

The spacecraft flies at a constant height, so that it can see a circular spot on the surface. It starts at the north pole and spirals down to the south pole, overlapping its previous track on each circuit. The alien cannot move fast enough to cross the track before the next circuit, so it is trapped inside a reducing area surrounding the south pole.

The value of $10$ is not critical, so we do not have to optimise the details. Take the height above the surface to be half the radius. Then a diameter of the spot subtends an angle $2 \cos^{-1}(1 / 1.5)$ at the center of the planet. $1 / 1.5 < 1 / \sqrt{2}$, so the angle is more than $90$ degrees. The critical case is evidently when the spacecraft is circling the equator. Using suitable units, we may take the radius of the planet to be $1$ and the spaceship speed to be $1$. Then the diameter of the spot is $\pi / 2$. We take the overlap to be $2 / 3$, so that each revolution the track advances $\pi / 6$. If the planet flew in a circle above the equator, the distance for a revolution would be $2\pi \times 1.5 = 3\pi$. The helical distance must be less than $3\pi + \pi / 6 = 19\pi / 6$. So the alien can travel a distance $19\pi / 60 < 2 / 3 \times \pi / 2$ and is thus trapped as claimed.