This is mostly just intuition, but it might be helpful. The "winnable" and "unwinnable" position definitions are based on the assumption that the player in the winnable position always makes a move that puts the other player in an unwinnable position. But, if you do not assume *a priori* that either player plays optimally, then you have to consider all of the possible moves in a winnable position, not just the best move.

For simplicity, I'll define a "winning" move as one that puts the opponent in an unwinnable position.

That gives us a way to compare winnable positions: one winnable position is better than another winnable position if a greater proportion of the available moves are winning moves. So, you could say that the optimal move in any "unwinnable" position is the move that puts the opponent in the worst available winnable position, i.e. maximizing the opponent's opportunity to blunder away the win.

We can easily generalize that to both players: the optimal move in any position is the one that gives the opponent the smallest proportion of winning moves out of all of their available moves. By definition, the winning player always has at least one move that gives the opponent zero winning moves, and those are exactly the optimal moves under the existing definition.

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If you have or can calculate a probability distribution over the opponent's moves, you can alternatively define optimal play as maximizing the probability of ever being in a winnable position, which can be calculated by recursively evaluating every subtree for every possible move. This again lines up with the existing optimal move definition: if a player is already in a winnable position, a winning move puts them in another winnable position with probability 1.