The intersection of a set $A \subset \mathbb{R^3}$ with all planes is always a circle. Prove that $A$ is a sphere.
We regard single points and $\emptyset$ as degenerate circles & spheres.
I was told a partial proof: we can assume that there's at least one non-degenerate intersection, because otherwise $A$ must be $\emptyset$ or a point (a sphere of radius $0$): indeed, if $x_1,x_2 \in A$ then we take a plane passing through both and arrive at a contradiction. Hence there is a circle $A \cap P =C \subset A$. Now we choose a diameter $D$ of $C$ and intersect $A$ with all planes that pass through D's two endpoints This should somehow imply that there's a sphere $S \subset A$, and then we note that if $A$ contains a sphere it cannot contain another any other point $x$.
I'd like an explanation of this argument or an alternative proof.
Take $C_{max}$ the circle of max diameter given by $A \cap P$, with $P$ a plane and $A$ the set.
There is at least one circle of max diameter. Take $D_{max}$ one of the diameter of $C_{max}$.
Take $P_{D_{max}}$ the family of planes that include $D_{max}$.
They all intersect $A$ forming a circle, of diameter already existing $D_{max}$.
By symmetry of rotation, this forms a sphere.