If we have $S$ a compact surface and a unitary vector $a \in R^3$ with a plan $\Pi_{a} \perp a$ such that $S$ is symmetric with respect to $\Pi_{a}$. How we can prove that $S$ is sphere?
We can use Alexander theorem to prove that.
Any advice about the solution structure please?
I am not sure what tools you have at your disposal, but here is how I would prove it. I don't like this proof, because it doesn't extend easily to the analogous statement in higher dimensions.
First, for a given unit vector $a$, there is exactly 1 such $\Pi_a$. For, if there are two, say $\Pi_a$ and $\Pi'_a$, then the composition of reflections about $\Pi_a$ and $\Pi'_a$ is a translation in the direction of $a$. By no compact surface can be translation invariant.
Moreover, the same argument shows that $\Pi_a = \Pi_{-a}$. It follows that, if $\rho_{a}$ denotes reflection across the plane $\Pi_a$, the map $f:\mathbb{R}P^2\rightarrow Iso(S)$ given by $f(a) = \rho_{a}$ is a homeomorphism onto its image. In particular, $Iso(S)$ contains a copy of $\mathbb{R}P^2$.
On the other hand, $Iso(S)$ is a compact Lie group (Meyers-Steenrod). Its dimension is at most $3$: the isotropy group at a point $p$, $G_p:=\{g\in G: g(p) = p\}$ is contained in $O(2)$, of dimensions $1$, and $\dim G - \dim G_p \leq \dim S$. It's dimension is also at least 3 because it contains $\mathbb{R}P^2$ and $\mathbb{R}P^2$ can't be given the structure of a Lie group (since, for example, it is non-orientable).
There are precisely three connected compact Lie groups of dimension 3: $T^3, SU(2)$ and $SO(3)$.
But $Iso(S)^0$, the identity component of $Iso(S)$ is not $T^3$ because $Iso(S)^0$ is not abelian. In fact, if $a,b$ are unit vectors with angle smaller than $\pi/2$ radians, then $\rho_a\circ \rho_b\neq \rho_b\circ \rho_a$.
So, $Iso(S)^0 = S^3$ or $SO(3)$. In any event, we will show that $S$ is isometric to $S^2$.
The action is non-trivial, so there is a point $p\in S$ for which $G_p\neq G$. Then the orbit through $p$ is diffeomorphic to $G/G_p$. The subgroups of $S^3$ and $SO(3)$ are known, and there are no subgroups of dimension $2$. In particular, $G_p$ has dimension at most $1$, so $G/G_p\subseteq S$ has dimension at least $2$. Since $S$ is $2$-dimensional and connected, $G/G_p = S$.
In particular, $S$ is a homogeneous space for $S^3$ or $SO(3)$. It follows that $S$ is isometric to $S^2$ or $\mathbb{R}P^2$ with usual round metric. But $\mathbb{R}P^2$ cannot be embedded into $\mathbb{R}^3$, so $S = S^2$ with round metric.