Here is my attempt:
If $A$ is nonsingular and $A=A^T$, then $A^{-1}A=A^{-1} A^T=I_n.$ Since $A^T$ is nonsingular, we multiply $(A^T)^{-1}$ on both sides and we get $(A^T)^{-1}I_n=A^{-1}I_n \implies (A^T)^{-1}=(A^{-1})^T=A^{-1}$. Thus $A^{-1}$ is symmetric.
In the instructor's manual, the proof goes like this:
We must show that $(A^{−1})^T = A^{−1}.$ First, $AA^{−1} = I_n$ implies that $(AA^{−1})^T = I_n^{T} = I_n$.
Now $(AA^{−1})^T = (A^{−1})^TA^T = (A^{−1})^{T}A$, which means that $(A^{−1})^T = A^{−1}$.
Is the author's approach more efficient, moreover, is my approach valid?