Because the analysis is performed assuming that there is a strict total order on the keys. Therefore if you are sorting, for example, integers on their usual order, the lower bound applies only as long as each key is different. If you allow the order to not be strict then the analysis is not particularly interesting, your example is indeed a best case.
Observe that the assumption of a strict total order is actually a small cheat, because we are also implicitly assuming that all comparisons cost $\Theta(1)$. If we disallow repeated keys, then the number of different keys is not bound by a constant, which means that the key size is also not bound by a constant, in contradiction with the constant cost assumption for comparisons.