as @Ricky Demer suggested, the restriction only shows you the NP-hardness, you also needs to show the problem belongs to NP in case to you want tho show the NP-completeness.
This answer is dedicated to answer your questions in your comment to Mr. Ricky's answer.
To show some problem is NP-hard, you need to prove worst case scenario is NP hard.
If you can show a NP hard problem is a special case for the problem you want to prove, then it that problem is also NP-hard. Because this special case showed the worst case is at least NP hard.
Of course there might be other cases in a NP hard problem that can be solved in polynomial time, but those are not the worst cases.
So regardless of the relation between P and NP, restriction is correct.
The technique you showed in your 2nd paragraph is not sufficient, Because as what I wrote in the previous paragraph, the problem you need to prove can be a case solvable in P as well, hence you cannot say it is NP-hard.
In addition, as long as you can find a proven NP hard special case for your problem, and the proof is easier than reduction or any other method, then it is a good idea to use restriction.
From RandomStudent: the reduction actually works in a similar way.
Have fun with cs :)