 Solution

An eigenvalue of $A$ is a scalar $\lambda$ for which $Az=\lambda z$ has a nonzero eigenvector. Allowing initially $z\in {\mathbb{C}}^m$, symmetry of the real matrix gives $A^{\ast }=A$, so $z^{\ast }Az$ is real. Since $z^{\ast }z>0$,This proves reality without presuming it when choosing the eigenvector. Once $\lambda$ is real, the real or imaginary part of $z$ supplies a nonzero real eigenvector.