Are non-trivial sets formed by set operations on NPC sets still in NPC?
Non-trivial sets formed from the unions and intersections of NP-complete sets are also not
necessarily NP-complete (of course this is all under the assumption
that P != NP).

Let $L$ be a NP-complete language, $A$ be any set in P, and $B:=
{\{0,1\}}^*$.  Then the following sets are all NP-complete since they
are just the disjoint unions of a sets based on $L$, $A$, and $B$.

$L_1 := 00L \cup 01A$

$L_2 := 11L \cup 01A$

$L_3 := 01L \cup 10B$

$L_4 := 10L \cup 01B$

Then $L_1 \cap L_2 = 01A$ and $L_3 \cup L_4 = 01B \cup 10B$
These results are both in P and and non-trivial sets.

On the other hand, as mentioned in a comment above, the Cartesian
product of two NP-complete sets is always NP-complete:  If $X$ and $Y$
are two NP-complete sets  then $X \times Y$ is in NP since a
NP-witness for $(x,y) \in X \times Y$ is simply a witness for $x \in
X$ along with a witness for $y \in Y$.  Furthermore, the map $x
\rightarrow (x,y_0)$ is a polynomial time reduction from $X$ to $X
\times Y$ where $y_0$ is a fixed member of $Y$.