Can sets $A,B,C$ satisfy $(A\in B \in C) \land (A \subseteq B \subseteq C)$? We have a union of criteria: $$(A\in B \in C) \land (A \subseteq B \subseteq C)$$
Can such sets exist? 
I spent some time thinking about this and I came up with this example: 
$$A = \{ a\}$$ $$B = \{ \{ a\}, a  \}$$ $$C = \{ \{  \{a \}, a\}, \{a\}, a \}$$ 
Can this actually work?
 A: Looks right to me!

It's worth noting that the key to the problem, which you picked up on at least intuitively but is worth saying out loud, is: sets of mixed rank. Intuitively, a set like "$\{a, \{a\}\}$" might feel weird since the elements "$a$" and "$\{a\}$" seem like different types of things, and outside of set theory we usually deal with sets of objects of the same type (e.g. sets of real numbers, sets of continuous functions, sets of groups, ...).
But the idea that sets of mixed rank are totally okay is a key intuition in set theory. Indeed, such sets are extremely useful - e.g. ordinals are such sets. This problem is (I think) trying to make you comfortable with such objects.

On the "ordinals" note above, note that $a$ isn't really playing any role in your solution: we could also have $$A=\{\}, B=\{\{\}\}, C=\{\{\}, \{\{\}\}\},$$ and these sets are exactly the first three ordinals (and this pattern continues: one of the key points about ordinals is that $\alpha\in\beta$ implies $\alpha\subseteq\beta$). Getting used to building stuff from the emptyset alone is another important part of learning set theory.
A: Perfect, you actually recovered the structure of the first nonzero ordinals: $1,2,3$. Starting with the set $a$ you applied the successor function $S(x):=x\cup\{x\}$ which is the first method for building bigger ordinals. Ordinals are $\in$-transitive. Id est $x\in y$ and $y\in z$ implies $x\in z$ which could seem strange at first glance, beating our level-idea of sets.
