Why must $A\subset B$ and $A \cap C = \varnothing$ for $A \cap (B-C) = A$ I really don't understand how I'm supposed to prove that $A \cap C = \varnothing$ . And why can't, for instance, B be a subset of A?
 A: Draw a diagram.
Here's a Venn diagram with $B\setminus C$ marked in green:

And then here's one with $A\cap(B\setminus C)$ marked in red:

If we're given that $A\cap(B\setminus C)$ equals $A$, then every element in $A$ must be in the red area. This area is completely outside $C$, so $A$ and $C$ must be disjoint, but it is completely inside $B$, so $A$ must be a subset of $B$.
$A$ doesn't have to be a strict subset of $B$, though -- it is possible under your assumptions that $A$ and $B$ are the same set.
A: So you want to show if $A \cap (B - C) = A$ then both $A \subseteq B$ and $A \cap C = \emptyset$.
Well, the first thing you have to realize is that $B - C = B \cap C^{c}$, where $C^{c}$ is the complement of $C$.  To convince yourself of this, draw two circles which overlap a little.  Call one of them $B$ and the other $C$.  $B - C$ is the part of $B$ that doesn't overlap with $C$.  Hopefully you'll see that this is the part of $B$ outside of $C$, i.e, in $C^{c}$, i.e., $B- C = B \cap C^{c}$.
Now, $A \cap (B - C) = A \cap (B \cap C^{c})$.
So we have $A \cap B \cap C^{c} = A$.  Notice that the left hand side of this equality is a subset of $C^{c}$.  Since this subset of $C^{c}$ equals $A$, that means $A \cap C = \emptyset$ (do you see why?).  
Also, $A \cap B \cap C^{c}$ is a subset of $B$.  So we have that this subset of $B$ equals $A$.  Of course that implies $A \subseteq B$.
A: Hint:
For any two sets $\;X,Y\;$ , we have that $\;X\cap Y=X\iff X\subset Y\;$ , so in your case you have that $\;A\subset B\setminus C\;$ ...
