Accepting the fact that, like all others normal matrices, a hermitian matrix (with $N$ lines) has a set of $N$ orthonormal complex eigenvectors (with real eigenvalues, degenerate or not), how do I prove that a symmetric matrix (real hermitian) has a set of $N$ orthonormal real eigenvectors?
It follows simply from the fact that the eigenvalues of the symmetric matrix are real. Assume that $A$ has a real eigenvalue $\lambda$ and a nonzero eigenvector $u=v+iw$, where $v$ and $w$ are real. Then it follows $Au=\lambda u$ that $Av=\lambda v$ and $Aw=\lambda w$ and hence $v$ and $w$ are in the same eigenspace associated with $\lambda$. Consequently, the eigenvectors can be chosen to be real.