So...where do you want the final methyl? You want one attached to the ring?
This is how I followed you:
Goal: You need to make B or C (which you can already do with 2 eq. of BuLi from A).
Attempts: Now you tried to convert A to C with 2 eq. of MeLi, but you get D
So then you tried to convert A to C with 1 eq of MeMgCl, which gave you E.
With E in hand, you tried to salvage it by adding one more MeMgCl, but you had no reaction.
You threw another potion at it
(this time MeLi), you had no reaction again.
So now you want to find a route to B or C, that avoids the fulvene moiety?
So I said, "ok, if I had D, C is just the 'H2' looking reduced form of D". So I was asking if there was something I'm missing about the reaction. Because I reasoned that the double bond that was not part of either aromatic moiety would be much more susceptible to reduction than the two rings which would then give you C, which is what you needed.
I believe sjb, was saying, you can make D directly through his reaction and then wittig the methyl group on, and then reduce, possibly?
How did I do on understanding the thread?