Let $\lambda_{1}$ and $\lambda_{2}$ be two distinct eigenvalues of matrix $A$ and $u$ and $v$ be their corresponding eigenvectors respectively.
We know that an eigenvector $X$ corresponding to an eigenvalue $\lambda$ of matrix $A$ satisfies
$AX = \lambda X$
$\therefore Au = \lambda _{1}u \quad \to (1)$ and $Av = \lambda _{2}v \quad\to (2)$
On pre-multiplying eqn $(1)$ with $v^{T}$, we get
$v^{T}Au = v^{T}\lambda _{1}u$
$\left ( v^{T}A\right )u = v^{T}\lambda _{1}u$
$\left ( A^{T}v\right )^{T}u = v^{T}\lambda _{1}u$
$\left ( Av\right )^{T}u = v^{T}\lambda _{1}u$ $($since $A$ is a symmetric matrix, we can write $A^{T} = A)$
But $Av = \lambda _{2}v$ ... from $(2)$
$\left ( \lambda _{2}v\right )^{T}u = v^{T}\lambda _{1}u$
$\lambda _{2}v^{T}u = v^{T}\lambda _{1}u$
$\lambda _{2}v^{T}u = \lambda _{1}v^{T}u$ $($as $\lambda _{1}$ is a constant, we can write $v^{T}\lambda _{1} = \lambda _{1}v^{T})$
$\lambda _{2}v^{T}u - \lambda _{1}v^{T}u = 0$
$\left ( \lambda _{2} - \lambda _{1} \right )v^{T}u = 0$
$\therefore$ Either $\left ( \lambda _{2} - \lambda _{1} \right ) = 0$ or $v^{T}u = 0$
But since $\lambda _{2} \neq \lambda _{1}$, $v^{T}u$ must be $0$
$v^{T}u$ is nothing but the dot product of the eigenvectors $u$ and $v$
Hence, we can conclude that the eigenvectors corresponding to distinct eigenvalues of a real symmetric matrix are orthogonal.