What I'm wondering is why solving those instances G resulting from reduction of 3-SAT to 3-COLOR is the same as solving all instances of 3-COLOR.
It's not. The point is to be able to solve just the 3-SAT examples.
If the 3-COLOR problem is about whether a graph with structure similar to G is 3-colorable, then this approach works. But clearly it isn't.
Absolutely. But the point of the proof is just to deal with those graphs, specifically.
It seems like here there's the implication that every instance of a particular problem is equally hard. At least here in this example, however, we know a lot about the graph G, and it seems to me G could be 3-colored easily and thus the corresponding 3-CNF could be solved easily. But again this doesn't say anything about other graphs that are not like G. Please explain this, and please also correct me here if I seemed to underestimate the difficulty of 3-coloring G.
3-colouring a general such $G$ is indeed hard. In fact, it is exactly as hard as solving 3-SAT!
This sentence in the link is key:
That is, given an instance φ of 3-SAT, we will construct an instance of 3-COLOURING (i.e. a graph G(V, E))
where G is 3-colourable iff φ is satisfiable.
As the second reason in your quote says, we do not care about creating an algorithm for solving 3-colouring. What we care about is showing that it is hard. We do that by showing that if 3-coloring was easy, then 3-SAT must also be easy. But we already know that 3-SAT is not easy. Thus, by contradiction, we must have that 3-coloring was not easy, i.e. it must be hard.
How can we do this? We need to do two things:
Show that any 3-SAT problem can be transformed into a 3-coloring problem in polynomial time. This equivalence is usually the harder part. In this case, one needs to show that, given some instance of a 3-SAT problem, we can get a graph s.t. that graph is 3-colorable if and only if the instance is satisfiable.
Show that if we could magically solve 3-coloring, then we could easily solve 3-SAT.
The second point basically automatically follows from the first, by just transforming a given instance from 3-SAT to 3-color. Do you see where the contradiction comes from? Since an equivalence exists, we can easily/quickly transform between 3-SAT and 3-color. Imagine we had an algorithm $A$ to quickly solve 3-color. But this is impossible, because then the following algorithm quickly solves 3-SAT: (1) transform from $\phi$ to $G$, and (2) use $A$ on $G$.
Notice that we do not need to actually come up with any coloring algorithms, or worry about general graphs and how to color them. We only need to establish a way to get a graph $G$ such that colorability becomes equivalent ("iff") to satisfiability of $\phi$. The proof is then by contradiction.
Note that it doesn't matter how easy or hard $G$ is to color or solve as a 3-color problem instance. It only matters that coloring $G$ tells us the answer to $\phi$, whether positive or negative.
TL;DR: we do not want to try to solve 3-color here, we just want to show we can transform $\phi$ into $G$. (This means if we could solve 3-color then we could solve 3-SAT, but we do not actually want an algorithm to do either of those here.) We can then prove by contradiction that 3-color is hard.
Edits for comments:
As I understand, the main point of the proof is to show that, as soon as we say we can solve G efficiently, we'd arrive at a contradiction. So an algorithm A to solve ALL instances of 3-COLOR can't exist. But this wouldn't exclude the possibility that there exists an algorithm to solve instances of 3-COLOR that are not isomorphic to G, right?
About this specific case, I'm not sure. Someone with more graph knowledge could maybe answer. Presumably you mean a poly-time algo. But note that we don't know if $P=NP$. If it does, then there exists an algorithm for all $G$ :)
As I understand, the statement for the 3-COLOR problem is:"Given any graph X, determine if it’s 3-colorable”. The process to arrive at the contradiction works if we can find a 3-CNF that can be reduced to X, or equivalently, if the graph G as defined above and corresponding to that 3-CNF is isomorphic to X. Only then an efficient way to answer that question for X would lead to a contradiction. Otherwise I just don't see how we can generalize the claim that there's no efficient way to answer the question for the set of graphs G to the set of all graphs X.
We don't care about all graphs $X$. We only care about $G$ that arise from 3-SAT. That is sufficient for the proof, since the hardness of those graphs alone is enough (since it is equivalent to 3-SAT). Other graphs don't matter.
The process to arrive at the contradiction works if we can find a 3-CNF that can be reduced to X, or equivalently, if the graph G as defined above and corresponding to that 3-CNF is isomorphic to X.
No. It's the other way around. The contradiction works if any 3-CNF can be reduced to some equivalent $G$. No other graphs need be considered. I.e., whether or not other graphs can be efficiently colored is unimportant here.
I just don't see how we can generalize the claim that there's no efficient way to answer the question for the set of graphs G to the set of all graphs X.
We don't need to generalize it. This fact alone is sufficient for the proof. Since these $G$ graphs are a subset of $X$, and just solving them alone is hard, clearly solving $X$ is equally hard or harder.