My move resulted in us both being in check. Is this legal?
No. Proof by contradiction. First we assume the premise that in the current position both White and Black are in check. Then we prove a contradiction (specifically that one player made an illegal move). Because the premise leads to a contradiction, the premise must be false; i.e., it cannot be the case that in a legal position both players are in check.
Assume for definitiveness (but without loss of generality) that Black is the player that made the last move that resulted in both players being in check.
Let P2 be the position that resulted from Black's move (i.e., in which both players are in check).
Let P1 be the position one half-move earlier, i.e., just prior to Black's last move. I.e., in P1 it is Black to move.
One of the two following statements must be true:
- In P1, Black is in check.
- In P1, Black is not in check.
For each of these two cases, we show that Black's move in P1 was illegal.
Case 1: Black is in check in both P1 and P2. In this case, Black's move was illegal because he was already in check but his move did not get him out of check. (Either (a) there was at least one legal move that would have allowed him to escape check but he didn't make that move or (b) he was in checkmate and the game should have ended instead of Black making a move.)
Case 2: Black is not in check in P1 but is in check in P2. In that case Black's last move is illegal because it resulted in him being in check (i.e., he "moved into check").
Thus, in every possible position P1 (with Black to move), it cannot be the case that a legal move by Black results in both players being in check simultaneously.
Of course, an analogous argument works if it is White who made the last move.