Proof that a closed and compact subset of plane having two axes of symmetry with angle of intersection not a rational multiple of $\pi$ is either a disc or a whole plane.
Proof The composite of two symmetries is a rotation through an angle that is not a rational multiple of $\pi$ about, the point of intersection. Thus the subset is invariant under rotations $\Phi_\alpha$ about that point through angles $\alpha$ forming an everywhere dense subset of the unit circle. Since the given subset is assumed to be closed, it must therefore be mapped to itself by every rotation about P. Hence the desired result folows from convexity.
My question: What difference does it make if the angle is rational multiple of $\pi$?
If you don't rule out rational multiples of $\pi$, you might instead have a regular polygon.
A regular $n$-gon has $n$ axes of symmetry, going though vertices and/or midpoints of sides (depends on whether $n$ is even or odd). Adjacent axes of symmetry meet at an angle of $2\pi/n$, so any two axes of symmetry meet at an angle that is a multiple of $2\pi /n$ (and, hence, a rational multiple of $\pi$).
(image taken from here)