Suppose $A$ is a symmetric matrix with complex entries. Let $c$ and $k$ be two distinct eigenvalues and $X$ and $Y$ be two eigenvectors belonging to $c$ and $k$. $AX=cX => (AX)^T=(cX)^T => X^TA^T=cX^T => X^TA=cX^T => X^TAY=cX^TY => X^TkY=cX^TY => (k-c)X^TY=0 => X^TY=0$,since $c≠k$.
My teacher gave the result for real symmetric matrices. I don't see why it won't be true for complex matrices. Isn't there a notion of orthogonality in $\Bbb C^n$?
Yes, there is. But the usual concept of orthogonal vectors in $\mathbb{C}^n$ is: $(z_1,\ldots,z_n)$ and $(w_1,\ldots,w_n)$ are orthogonal when $\sum_{j=1}^nz_j\overline{w_j}=0$. And, for this concept of orthogonal vectors, it is not true in general that eigenvectors belonging to distinct eigenvalues of a complex symmetric matrix are orthogonal.
However, it is true for Hermitian matrices (and the proof is similar).