If T is auctioned on the 3d and L is on the fourth, it means that H must be auctioned on the first or the second because S cannot be auctioned first, and S must be auctioned before V, thus V cannot be second and V must be auctioned after T. H thus must be auctioned first and S must be auctioned second since S cannot be #5 or #6, but then we have no space left for M, which must also be auctioned before L per rule 2, thus (D) must be false.

H S T L

(B) is correct because we could come up with a scenario that complies with all the rules where S is 4th: