Does a reduction from sorting imply an $\Omega(n \log n)$ time bound?
It's not a valid argument.   What it does prove is that any comparison-based algorithm for Problem X must use at least $\Omega(n \log n)$ comparisons.  (There is a known algorithm for integer sorting whose running time is $O(n \sqrt{\log \log n})$, in some models of computation, but it is not a purely comparison-based algorithm; see here.)