On the Wikipedia page for strategy-stealing arguments, there is an example of such an argument applied to tic-tac-toe:
A strategy-stealing argument for tic-tac-toe goes like this: suppose that the second player has a guaranteed winning strategy, which we will call S. We can convert S into a winning strategy for the first player. The first player should make his first move at random; thereafter he should pretend to be the second player, "stealing" the second player's strategy S, and follow strategy S, which by hypothesis will result in a victory for him. If strategy S calls for him to move in a square that he has already moved in, he should choose at random again. This will not interfere with the execution of S, and this strategy is always at least as good as S since having an extra marked square on the board is never a disadvantage in tic-tac-toe.
My confusion is about the part in bold. I understand the second part - if you have more squares marked, you will always do at least as well as a setup where you have fewer squares marked - but I don't understand why this won't interfere with the optimal strategy.
Can someone explain why being forced to make a move that you've already made will not interfere with a winning strategy?