Here is a "geometric" argument for the real case. The skew-symmetric condition is equivalent to $\langle Av, v\rangle =0$ for all vectors $v$. Geometrically, $Av$ is orthogonal to $v$. If $Av$ is never zero for $v\ne 0$, then we get a nonvanishing vector field on the unit sphere, contradicting the Hairy Ball theorem.
Notice that $\langle A, B \rangle = \sum_{i,j} [A]_{ij} [B]_{ij}$. Let $E_{ij} = e_i e_j^T$, ie, the zero matrix except for a one in the $(i,j)$ position.
Suppose $\langle A, S \rangle = 0$ for all symmetric matrices, then it is true for $S=E_{ij}+E_{ji}$. This gives $\langle A, E_{ij} \rangle + \langle A, E_{ji} \rangle = 0$, which gives $[A]_{ij} + [A]_{ji} = 0$, from which it follows that $A = -A^T$.
Now suppose that $A$ is skew symmetric, and $S$ is symmetric. Then we can write $S = U^T + \Lambda + U$, where $\Lambda$ is diagonal, and $U$ is strictly upper triangular. Then $\langle A, \Lambda \rangle = 0$ because $[A]_{ii} = 0$. Since $\operatorname{tr} M = \operatorname{tr} M^T$ we also have $\langle A, U^T \rangle = \langle A^T, U \rangle$, and skew symmetry gives $\langle A^T, U \rangle = - \langle A, U \rangle$. Hence $\langle A, S \rangle = \langle A, U^T \rangle + \langle A, \Lambda \rangle - \langle A, U^T \rangle = 0$.
Best Answer
Hint: for two real vectors $v,w$, we note that $$ \langle v,w \rangle = v^Tw $$ Now, consider what this means when $$ v = Ax\\ w = y $$ Noting that for multiplicatively compatible matrices $A,B$, we have $$ (AB)^T=B^TA^T $$