Prove that $S$ is a sphere. Let $S\subset {\mathbb{R}}^3$ with the following properties:
$1.$ For any line $l$:
$|l\cap S|=2$ or $|l\cap S|=1$ or $|l\cap S|=0$
$2.$  For any plane $P$
$P\cap S=\text{circle}$ or $|P\cap S|=1$ or $|P\cap S|=0$
Prove that $S$ is a sphere.
 A: Assume $S$ contains more than one points. Let's say $p_1, p_2 \in S$. Let $P$ be any plane containing $p_1$ and $p_2$, then $C = S \cap P$ is a circle. Choose a coordinate system
such that $C$ becomes the unit circle on the $xy$-plane. i.e.
$$C  = S \cap P = \{\;(x,y,z) \in \mathbb{R}^3 : x^2+y^2 = 1, z = 0\;\}$$
Now consider the intersection of $S$ with the $xz$-plane:
$$C' = S \cap \{\;(x,y,z) \in \mathbb{R}^3 : y = 0\;\}$$
$C'$ contains two points $( \pm 1, 0, 0 )$ from $C$, so $C'$ itself is a circle.
It is clear $C'$ intersect the $z$-axis in two points. Let $u_1 = (0,0,d_1)$ and $u_2 = (0,0,d_2)$ be the two intersections and $v = (0,0,\frac{d_1+d_2}{2})$ be the center of $C'$.
Now consider the collection of planes $P_{\theta} : \theta \in [0,\pi)$ containing the $z$ axis. $P_{\theta}$ is obtained from the $xz$-plane by rotating it with respect to the $z$-axis for an angle $\theta$. It is clear $P_{\theta} \cap S$ contains at least 4 points from $S$:
$$u_i = (0,0,d_i)\quad\text{ and }\quad \pm (\cos\theta,\sin\theta,0) \in C$$
So $P_\theta \cap S$ is again a circle. Since three points on a plane determine the
circle and using symmetry, we see $P_\theta \cap S$ share the same center $v$ with $P_0 \cap S$. When $\theta$ varies from $0$ to $\pi$, the circles $P_\theta \cap S$ trace 
out a sphere centered at $v$.
Conclusion: If $S$ contains more than one points, then it is a non-degenerate sphere.
Thoughts
The above discussion is interpreting the circle as a geometric circle in Euclidean geometry. Let's see what can we salvage if we reinterpret the word circle in topological
setting. i.e replace the concept circle by a simple Jordan curve.
The existence of $C$ and $C'$ doesn't seems to have any issue. By a suitable choice of coordinate, we can still assume $C$ and $C'$ contains $(\pm 1,0,0)$ and $C'$ continue to intersect the $z$-axis at two points $(0,0,d_1), (0,0,d_2)$ because the $z$-axis split the $xz$-plane in two components. It is also clear the point $v = (0,0,\frac{d_1+d_2}{2})$ lies in the bounded component in the $xz$-plane associated with the Jordan curve $C'$.
One consequence of this is any ray in $xz$-plane starting at $v$ intersect $C'$ and hence $S$ at one and only one point. As we change $\theta$ and following the intersection
of $P_\theta$ with $S$, this will set up a bijection $\varphi$ between the unit sphere $S^2$ (direction of rays starting at $v$) and points on $S$. $\varphi^{-1}$ is clearly
continuous. It seems to me $\varphi$ is also continuous but without further assumption,
I can't justify this part. 
To me, $S$ seems to be a topological sphere but I don't have any idea how to prove that.
