It turns out that we can currently prove very little about the Diamonds problem. But here is the proof that we can give.
Theorem: If there are an odd number of slots, A will win.
Proof: This mirrors the result of the 5 and 7 slot games.
If the number of slots is odd, then A should take the centre slot. Whatever B does next will either give A the chance to put 3 in a row or it will not. If it does A will win. If it doesn’t then suppose that B’s move is on the left of centre. Then, since only two diamonds are on the board, there must be a ‘safe’ vacant slot on the right of centre where A can play. Clearly if B’s move is on the right of centre A plays symmetrically opposite on the left.
Now we can assume that the game has been progressing for some time. To this point, B and then A have occupied symmetric slots and the unoccupied slots also come in symmetric pairs. It is now B’s turn. Either B has a safe move or she does not. In the second case A wins. In the first case A occupies a slot that is symmetric with B’s last move.
Eventually B must run out of safe slots as the number of slots is finite. Then A wins.
The proof here is actually a proof by Mathematical Induction. If you would like to use a more formal approach than I have done you might look up Mathematical Induction on the web and use it with the above theorem.