NP-Completeness - Proof by Restriction
There is an implicit (trivial) reduction, from each instance of the restricted version to itself, really just relabelling it as an instance of the unrestricted problem.

To give an example that I think is more illuminating than Partition and Multiprocessor Scheduling, we can look at Vertex Cover and Hitting Set:


  Vertex Cover
  Input: A graph $G=(V,E)$, a positive integer $k$.
  Question: Is there a set $V'\subset V$ of size at most $k$ such that for every $uv \in E$ either $u\in V'$ or $v\in V'$ or both?
  
  Hitting Set
  Input: A hypergraph $H=(X,E)$, a positive integer $k$.
  Question: Is there a subset $X'\subseteq X$ of size at most $k$ such that for every $e \in E$ there exists an $x \in X'$ such that $x \in e$?


So we can easily see that Vertex Cover is just Hitting Set where the size of the edges is exactly two. As Vertex Cover is NP-hard (indeed, complete), Hitting Set must also be NP-hard. As Hitting Set is also in NP, it is therefore NP-complete.

So we have a "reduction" from Vertex Cover to Hitting Set, but it's just the identity mapping. The graph remains unchanged (as does $k$), we just call it a hypergraph, and call the edges hyperedges.

To think about it from the other end, if we have a problem $\Pi$ where a restricted subproblem $\Pi'$ is already NP-hard, and we had a PTIME algorithm for $\Pi$, then we could use it to solve all the instances of $\Pi'$ in polynomial time and hence everything in NP would have a polynomial time algorithm. Thus $\Pi$ must be NP-hard (and if $\Pi \in$ NP, NP-complete).