If $A$ reduces to $B$ and $B$ is NP-hard, is $A$ NP-hard?
B being NP-hard does not say anything about A. The problem says that there is a way to reduce A to B in polynomial time.
But on the other hand, it is also not given that A getting reduced to B is the only pathway for solving A. So, there can be other ways of solving A in polynomial time.
Hence, the implication that A is NP-hard is false.