Express unit sphere as countable union of great circles? 
Let $S = \{x\in \mathbb{R^3} | d(x,(0,0,0))=1\}.$   Is it possible that $S$ is a countable union of “great circles”? A great circle is the intersection of $S$ with a plane through $(0,0,0)$.

What I know is that the sphere is closed and bounded, so it is compact, so given any open cover there is a subcover. But great circles are closed, so probably it's not gonna work out.
Then I thought of using contradiction. But each great circles are uncountable. Countably union of uncountable set is still uncountable.
Any hint?
 A: you may reduce the dimension of your question by considering the intersections of your great circles with a suitably chosen equator.
so can a circle be made up of a countable set of points?
A: If you have measure theory available, you can recognize that great circles have zero area and so any countable union of them will have zero area, but the sphere has positive area.
A: Hint: the Baire category theorem.
A: To elaborate on the answer hinting at Baire Category:
The sphere $S^2$ is a complete metric space (use metric induced by Euclidean metric restricted to $S^2$). By Baire, a complete metric space cannot be a countable union of nowhere dense sets. But a "great circle" is nowhere dense: It is closed and its interior is empty (both easy to see).
A: suppose first that every great circle is in the collection. then the sphere is also covered by the poles of the great circles. but there are only 2 poles per great circle, so the sphere itself has only a countable set of points. contradiction.
so there is a great circle not included in the collection. but this is now the union of its intersections with the circles of the collection, and we obtain a contradiction as before. this time because any two great circles meet in exactly  two points.
