The decision version of the 3-COLOR problem is the problem of deciding whether an input graph G(V, E) can be colored using only 3 colors so that no 2 adjacent vertices have the same color.
I had interpreted that to mean that we were looking for any coloring.
However, most proofs I have seen that reduce 3-SAT to 3-COLOR to prove that 3-SAT is NP-Complete use subgraph "gadgets" where some of the nodes are already colored.
But in this case, it would only show that a specific 3-coloring (i.e. some nodes on the input graph are pre-colored) does not exist.
It doesn't show that no 3-coloring exists. Can someone clarify why the reduction is valid?