Just a quick check: I'm not very good at proofs so I was seeing if this was correct:
Let $A$ be a symmetric $n \times n$ matrix and let $\lambda_1,\lambda_2$ be the eigenvalues corresponding to the eigenvectors $v_1,v_2$ respectively. (Previously shown that all eigenvalues are real).
Then we want to show that $\langle v_1 | v_2 \rangle = 0$.
Consider $$\langle v_1 | v_2 \rangle = \langle \frac{Av_1}{\lambda_1} | \frac{Av_2}{\lambda_2}\rangle \\ \\ = \frac{1}{\lambda_1 \lambda_2}\langle Av_1|Av_2 \rangle \\ = \frac{1}{\lambda_1 \lambda_2}\langle v_1|A^{\star}Av_2 \rangle \\ = \frac{1}{\lambda_1 \lambda_2}\langle v_1|A\lambda_2 v_2 \rangle \\=\frac{1}{\lambda_1 \lambda_2}\langle v_1|{\lambda_2}^2 v_2 \rangle\\ =\frac{\lambda_2}{\lambda_1 }\langle v_1|v_2 \rangle$$
and so we have
$$\lambda_1 \langle v_1 | v_2 \rangle = \lambda_2 \langle v_1 | v_2\rangle$$
hence the result follows as $\lambda_1 \neq \lambda_2$.
I'm not sure if this would cover the case when one of the eigenvalues are zero though....