Characterization of the sphere Hello I wanted to prove the following statement. 

Let $M$ be a compact connected surface in $\mathbb{R}^3$ such that for all $d\in S^2$ there exists a plane called $\pi_d$, such that is orthogonal to $d$ and it's a plane of symmetry of $M$, then $M=S^2$.

So far, I could prove that if the origin is contained in every normal line to $M$, then $M$ is a sphere centered at the origin. So now I tried to see that if I have the symmetry plane given by the statement, every such plane contains the origin, and this way I think I could conclude, but I am not able to prove this last assertion.
 A: Let $n\in S^2$ and $\alpha$ and angle. Choose two vectors $n_1,n_2\in S^2$ with $n_1\times n_2=n$ and $2\arccos\langle n_1,n_2\rangle=\alpha$. Then the consecutive reflection at the planes $\pi_{n_1}$ and $\pi_{n_2}$ will give a rotation around $n$ with angle $\alpha$. As the used reflections are symmetries, so is the rotation. A surface for which any rotation is a symmetry must be a sphere.
A: Like you mentioned, it's all about showing that the different planes of symmetry all pass through the same point. Take three symmetries $\sigma_1$, $\sigma_2$, $\sigma_3$in pairwise perpedicular planes, interesecting at $0$. Let another plane of symmetry be $\pi$. Now check that for a symmetry $\sigma$  ( in fact, any isometric $\pi$) and another symmetry $S_{\pi}$ with respect to the plane $\pi$ we have
$$\sigma \circ S_{\pi} \circ \sigma^{-1} = S_{\sigma(\pi)}$$ the reflection w r to the plane $\sigma(\pi)$.
Since $\sigma_1$, $\sigma_2$, $\sigma_3$ and $S_{\pi}$ invariante $M$ we conclude that $S_{\sigma_1 \sigma_2 \sigma_3(\pi)}$ also invariante it. But $\pi' \colon = \sigma_1 \sigma_2 \sigma_3 (\pi)$ is the symmetric of $\pi$ with respect to the point $0$. It is a plane parallel to $\pi$, and, if $\pi$ does not contain $0$, distinct from $\pi$. But then also $S_{\pi'} \circ \S_{\pi}$ invariantes $M$, and this is a translation by a $\ne 0$ vector, not possible. 
Obs: $M$ matters just a bit, rather the whole thing is about the group generated by these reflection. Since it cannot   contain not trivial  translation, it will be the group of isometries fixing a certain point $0$ ( the fact that the reflections generate that is rather standard. 
A: Assume that $M$ is 2-dimensional orientable closed surface in
$\mathbb{R}^3$.
(1) For $x\in (M,\rho)$, where $\rho$ is a metric in $M$, we have a
plane $P$ and a reflection $r_P$ wrt the plane $P$.
So we have a claim that $M\cap P$ is an image of geodesic path.
Proof : Assume that injectivity radius of $x$ is
 $\epsilon$ and $\rho(x,y)<\varepsilon $.
If a shortest path $c$ between $x$ and $y$ is not in $M\cap P$, then $r_P(c)\neq c$
 is a shortest path between $x$ and $y$. It is a contradiction.
(2) Similarly, for $x'\in M$ we have a plane $P'$.
If $M\cap P,\ M\cap P'$ do not intersect, then by using $r_P,\
r_{P'}$, there is infinite isometric copies in $M$ of $M\cap P,\
M\cap P'$ s.t. they do not intersect pairwisely. Hence $M$ has an
infinite diameter. It is a contradiction.
Hence $M\cap P,\ M\cap P'$ intersect at $x$ with an angle $t$. Here
a composition of $r_P,\ r_{P'}$ is a $2t$-rotation (M. Winter
commented).
(3) Consider another point $x''$ so that we have a plane $P''$. That
is from $M\cap P,\ M\cap P',\ M\cap P''$ we have a geodesic triangle
$\Delta$. In further, at each vertex we have rotation isometry and
at each side we have reflection isometry. From these isometries, we
consider all copies of $\Delta$ so that they cover $M$. They may
overlap.
If some two distinct copies overlap in nonzero measure, then there is a geodesic triangle $\Delta' $, which is the overlapping.
 Hence we consider a cover of copies of $\Delta'$.
If any two distinct copies do not overlap in nonzero measure, then
 consider another point $x'''$ so that we obtain $\Delta ''$ which has smaller area than that of
 $\Delta$.
That is, $M$ has a cover of copies of a geodesic triangle $\Delta$
 where area of $\Delta$ can be arbitrary small. If $x\in \Delta$, then any point $y$ in $M$ has a point $y'$ with small $
 \rho(y,y')$ s.t. $x,\ y'$ has same curvature. Hence $M$ is a sphere of constant curvature.
