In case this helps:
The members of A are 1 and 2.
The subsets of A are {}, {1}, {2}, and {1, 2}.
The members of C are {1} and {1, 2}.
The subsets of C are {}, {{1}}, {{1, 2}}, and {{1}, {1, 2}}.
C is a set of sets, so every subset of C must also be a set of sets (i.e. a set consisting of all, some, or none of the sets that make up C).
If C were a set of chickens, every subset of C would also have to be a set of chickens (and every member of C would have to be a chicken).
Maybe to put it a different way, consider the set:
D = {1, 2, {1, 2} }
Then A ⊂ C and A ∈ C are both true. The former because D contains 1 and 2, which are all of the elements of A. The later because D contains {1, 2} which is A itself.
The important thing is that “is an element of” does not “propagate upwards”. The statement “S contains a set that contains x” is different from “S contains x”.
Thanks, guys! Slowly but surely, I think it’s getting into my ossified brain.
Indistinguishable, if I may extend your SDMB analogy - A is not a subset of C, because one has to follow the two click rule to get from C to the elements of A, right?
Thudlow, if I understand correctly, if a set contains a set (as in {1,2} in C), it doesn’t contain the elements within that set.
Leachim, I think you meant D rather than C, but setting that aside, I must be missing something: I get that A ⊂ D because the elements 1 and 2 are in both A and D, but it seems from the answers in Apostol that the set {1,2} wouldn’t apply, which is what’s confusing me in the first place. (Which is basically the situation in the question I posted in post 53.)
Including the set of clean-shaven barbers in a town where the there is only one barber and he shaves everyone who doesn’t shave themselves?
Correct—unless you include them separately, as leahcim did with his set D.
I’m not quite sure what you mean here, but let me take one more crack at your original question:
By definition, for A to be a subset of C means that every member of A is also a member of C.
Thus, if you can find a member of A that is not also a member of C, A is not a subset of C.
In your example, the members of A are 1 and 2. The members of C are {1} and {1,2}. So, none of the members of A are also members of C. As leahcim said, “is an element of” does not propagate upwards. If one set is a member of a second set, that doesn’t automatically make the members of the first set members of the second set also.
If you want to really blow your mind with sets of sets of sets, check out the Von Neumann construction of the natural numbers.
Yes, I did mean D, rather than C.
I don’t know what you mean by “the set {1,2} wouldn’t apply”? You mean be an element of?
Thudlow - Gotcha ya!
Leachim - I just realized I misread your post, so I’m all set! 
Right.
One last question: we’ve established that set E can be a member of F without being a subset of it. But does the reverse hold true? If E is indeed a subset of F, does that mean it is a member of F as well?
No. Members and subsets are different sorts of things, and it’s quite rare for E to be both a subset of F and a member of F, although leahcim contrived an example where it did happen.
See, there’s the concept I was missing! I somehow had it in my head that being a member of a set was a particular form of being a subset. Or vice versa. Or something. But to quote Jules, “hell, they ain’t even the same sport.”
You all fought some ignorance today! Thank you.
Raymond Smullyan, in What Is The Name Of This Book (you can find the full text on-line), resolves this paradox, or others like it. His resolution: There is no such thus-described town and can’t be any such thus-described town because the given description of the town is self-contradictory. Therefore, the person who is telling you about this town is just outright lying (or mistaken) about it.
He gives a simpler stripped down example: In a certain town, there are two residents named Jack and Jill. Jack is taller than Jill, and Jill is taller than Jack. How do you explain that? The explanation: Smullyan, or whoever is telling you this, is simply lying.
What about Omniscient, our original OP? Whatever became of his project?