In proof by contradiction, we assume a statement S is false, and thus $\lnot S$ is true, and then show $\lnot S$ implies some contradiction. This is normally show in the predicate logic as $( T \cup \lnot S \vdash \bot ) \to (T \vdash S)$, where T is any first order theory and $\bot$ is a contradiction. My question is related to the truth value of $\lnot S$. There could exist some model where in $(T \cup \lnot S)$ $\lnot S$ is false, there is a valid sequence of steps to $\bot$, and $\bot$ is true. In fact, $\lnot S$ could be false in all models.
Is it a requirement for a proof by contradiction that the added statement $\lnot S$ must also be true in all models for the proof by contradiction to be valid and sound?
If we can add a false statement, how does one prove $\bot$ when we can never show the proof is sound (for all models)?