Here is my (WRONG!) proposal:
THIS PROOF COMPRISES AN ERROR (cf. Roberto, below).

Better try this one:

I think your proof of "a_[rs a_u]v implies that a_rs is simple " is not correct.

The proof is based on the assumption that there is a vector x^v satisfying condition
x^m x^n a_mn =1

This is never possible, because being a_mn antisymmetric and x^m x^n symmetric their contraction is always 0.
This can easily be seen in the simple 2 dimensional case where
a_11=a_22=0, a_12 = -a_21=a
then
x^m x^n a_mn= x_1 x_2 a - x_2 x_1 a = 0.

The formal proof for the general case is:
x^m x^n a_mn = x^n x^m a_nm = - x^n x^m a_mn = - x^m x^n a_mn
(where in the first step I just renamed the contracted indices and in second step antisymmetry of a was exploited) therefore x^m x^n a_mn =0.

You are right, Roberto! Thanks for the hint and sorry for the error.
I hope this fixes the bug: