3-SAT is at least as hard as SAT, or SAT ≤p 3-SAT. We know this because we can translate arbitrary SAT instances into 3-SAT in polynomial time. 3-SAT is in NP because we can check a candidate solution in polynomial time by plugging in the values and evaluating.
More accurate than “an NP-complete subset of SAT” would be to say that SAT is polynomial-time reducible to 3-SAT or that we can solve SAT in terms of 3-SAT.
Combining both of these, each is reducible to the other. In some sense, they are the same problem. 3-SAT imposes just enough structure to force it into NP-complete. 2-SAT is in P.
