What is a more difficult problem which involves using the king to prevent castling than the following problem?
White to play and win:
[FEN "r3k3/5p2/2K2P1P/b1N1P2P/8/8/4n3/8 w q - 0 1"]
In this famous problem by H. Hultberg (1944), the white king castles to prevent Black from castling:
[FEN "r3k2/1p1p2p1/2pP3p/8/8/5R2/PPPP4/4K2R - - - 0 0 "]
White to mate in two moves.
According to chess problem conventions, castling is presumed to be legal unless it's provably illegal. In this position, you can prove that at least one player has lost the right to castle, but you don't know which one. (If the white rook on f3 came from a1, the white king must have moved to let it out. If the rook on f3 is a promoted piece, it must have visited a8, e8, or f8 before escaping the 8th rank, so the black king or rook must have moved.)
1.Rhf1? fails to 1.0-0-0.
1.0-0! and 2.Rf8#, as Black is now unable to castle.
In the comments section of the accepted answer, the OP (I frankly don't care that they're account is gone!) asked for a castling preventing castling problem where both castle rights are known, instead of the use retrograde analysis that that answer utilizes.
As such, out of plain fun and curiosity, I have made such a problem. It can be seen in the below diagram.
[Title "White To Move And Mate In 4 Moves"] [FEN "4k2r/2p1p2p/P1P1P2P/8/8/8/6P1/nn2K2R w Kk - 0 1"]
Clearly, in order to checkmate Black in just four moves, White's pawn must be promoted. However, if White pushes their pawn, then Black will castle and their is no mate! Thus, White must prevent Black from castling The only way to do that is with their rook.
But if White plays 1. Rf1?, then Black will play 1... Nc2+!, also preventing a mate. Therefore, White's only option to prevent Black from castling is to castle themself-1. 0-0.
Black now has three possible lines of defense that all consist of giving a check. will do a small briefing on them here:
-1... Nd2 2. a7 Nf3+ 3. Rxf3 (gxf3? 0-0!) ~ 4. a8=Q/R#
-1... Rg8 2. a7 Rxg2+ 3. Kxg2 ~ 4. a8=Q/R#
-1... Rf8 2. Rxf8+ (a7? Rxf1+ 3. Kxf1 Nd2+!) Kxf8 3. a7 ~ 4. a8=Q/R#
And there we have it, a problem where both castling rights are known and a castling to prevent castling is the only way to win!
A 3-mover version just for fun. I will not give the solution though.
[FEN "4k2r/4p1p1/1P2P3/1N6/8/8/7P/n3K2R w Kk - 0 1"]