If $A$ is a complex $n \times n$ matrix such that $A^t = - A$, then $A$ is $0$.
What's wrong with the following proof?
Let $J$ be the Jordan form of $A$. Since $A^t = -A, J^t = -J$. But $J$ is triangular so that $J^T = —J$ implies that every entry of $J$ is zero. Since $J = 0$ and $A$ is similar to $J$, we see that $A = 0$.
I think that $A = P^{-1} J P \implies A^t = P^t J^t (P^{-1})^t$ and $A^t = -A \implies P^t J^t (P^{-1})^t = -P^{-1} J P$ from here can we always conclude that $J^t = -J$ as $P$ is not always orthogonal.
Ref: Hoffman Kunze Sec 7.3 No 14.
The claim is false:
$$A=\begin{pmatrix}0&\!-1\\1&0\end{pmatrix}\implies A^t=\begin{pmatrix}0&1\\\!-1&0\end{pmatrix}=-A$$
The part that is wrong in the (rather weird, imo) argument with the JCF, is that you didn't explained why from $\;J\;$ being triangular it follows that $\;J^t=-J\implies J=0\;$ ...I don't see it clearly at all in your proof (it is true, though).
You also didn't explain why the JCF of $\;A\;$ would have to fulfill the same condition as $\;A\;$ (this isn't true and the above is a counter example)