You are needing to assume that the eigenvalues are different, no? Otherwise, $A=0$ satisfies $A=A^\top$ and $Ax=0x$ as well as $A2x=0(2x)$ yet $\langle x,2x\rangle\ne0$. So, basically the idea then comes from the fact that $A^\top$ is the adjoint for $A$ on $\mathbb{R}^n$ with the usual inner product. In other words, $A^\top$ satisfies the relation $\langle Ax,y\rangle=\langle x,A^\top y\rangle$. So, since $\lambda_1\ne\lambda_2$ we may assume WLOG that $\lambda_1\ne0$. In this case we have that
$\begin{aligned}\langle v_1,v_2\rangle &=\frac{1}{\lambda_1}\langle \lambda_1 v_1,v_2\rangle\\ &=\frac{1}{\lambda_1}\langle Av_1,v_2\rangle\\ &=\frac{1}{\lambda_1}\langle v_1,A^\top v_2\rangle\\ &=\frac{1}{\lambda_1}\langle v_1,Av_2\rangle\\ &=\frac{1}{\lambda_1}\langle v_1,\lambda_2 v_2\rangle\\\ &=\frac{\lambda_2}{\lambda_1}\langle v_1,v_2\rangle\end{aligned}$
and since $\lambda_2\ne\lambda_1$ this implies that $\langle v_1,v_2\rangle=0$.