Proof or find a counterexample:For all sets $A;B;C$ if $A\subseteq B,\ B\subseteq C,$ and $C\subseteq A,$ then $A=B=C.$ Proof or find a counterexample:For all sets $A;B;C$ if $A\subseteq B,\ B\subseteq C,$ and $C\subseteq A,$ then $A=B=C.$
My solution:
True. Let $x\in A$, and since $A\subseteq B$
  this implies that $x\in B$
  and since $B\subseteq C$
  this implies that $x\in C$. Thus, 
$A\subseteq C$. Since $C\subseteq A$, 
then $A=C$. Then let $y\in C$
  and since $C\subseteq B$
  this implies that $y\in B$. And since $C\subseteq A$
  this implies that $y\in A$. Thus, 
$A\subseteq B$. Since $B\subseteq A$, then $A=B$. Hence this proves that $A=B=C$.
Can I get feedback on my answer?
 A: Since the three subset relations go in a circle among the three sets, the heart of the proof should be to show that some pair of those sets are equal (for example, $A=C$). But you don't want to have to repeat the same proof steps two more times; it's tiresome and can lead to slip-ups.  In fact your first slip-up is immediately after the statement $A=C$: "let $y \in C$ and since $C \subseteq B$" etc.  In fact,  $C \subseteq B$ was what you might have proved (but had not yet proved) by an argument that started with arbitrary selection of  $y \in C$.
One approach might be, once you have proved $A=C$, invoke the fact that by relabeling the three sets, the same premises also result in $A=B$ (or $B=C$, depending on how you do the relabeling). Another way to frame this is to say that you will prove that no matter which two of the three sets are selected, those two sets are equal, and without loss of generality you will assume the two sets arbitrarily picked are $A$ and $C$.
Another way to apply the same principle starts with different symbols altogether.  You could put it this way: for any three sets $X$, $Y$, $Z$, if $X \subseteq Y$, $Y \subseteq Z$, and $Z \subseteq X$, then (using the same method you used for $A = C$) you can show that $X = Z$.  From this and your original premises it follows immediately that $A=C$. (Consider how the $X=Z$ result applies when $X$, $Y$, and $Z$ are $A$, $B$, and $C$ respectively.)  But it also follows just as easily that $A=B$ (consider the case where $X$, $Y$, and $Z$ are $A$, $C$, and $B$ respectively), so $A=B=C$.
