Is my theorem about $P \neq NP$ correct?
I don't think the theorem is correct.  At least, your proof is not correct.  The problem is in the second sentences of your proof:


  Since the problem is NP-complete, any NP problem can be reduced to it in $O(N^r)$ steps, for some $r$.


This statement is not correct.  There's no guarantee there is a single $r$ that works for every NP problem.  You might have a different value of $r$ for each NP problem, with no upper bound on the set of all $r$.  (To give a silly example: Maybe for 3SAT $r=3$, for 4SAT $r=4$, for 5SAT $r=5$, and so on.  Don't take this example too seriously; it's just to give you the idea.)