Why every language in np is polynomial time reducible

Question

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Q: Show the error in the following fallacious "proof" that P ≠NP. Proof by contradiction. Assume that P = NP. Then SAT ∈ P. So, of some k, SAT ∈ TIME(n^k). Because every language in NP is polynomial time reducible to SAT, NP ⊆ TIME(n^k). Therefore P ⊆ TIME(n^k). But, by the time hierarchy theorem, TIME(n^k+1) contains a language which isn't in TIME(nk), which contradicts P ⊆ TIME(n^k). Therefore P≠ NP.