Is my proof correct? Prove that $A \subseteq B \implies C-B \subseteq C-A$. Problem: Prove that $A \subseteq B \implies C-B \subseteq C-A$.
My attempt: 
$A \subseteq B \implies$ if $x \in A$ then $x \in B$. Since $A$ consists only of some (or all) elements in $B$, we remove less elements of $C$ when taking the set difference, $C - A$, than we do when taking the set difference, $C - B$. Therefore, since for both of these differences we are removing elements from the same set, $C - B \subseteq C-A$.
I appreciate this is pretty sloppy, but I think the logic is almost there, I'm just not sure how to formalise the ideas.
 A: Formally:
$A\subset B$ $\rightarrow $
$B^c\subset A^c$ $\rightarrow$
$C\cap B^c  \subset C\cap A^c$.
With
$C - B = C\cap B^c;$  $C-A = C\cap A^c$,
we get:
$C-B \subset C-A.$
Appended: 
Proof of $A\subset B$  $\rightarrow$  $B^c\subset A^c$.
$A\subset B $:   $x\in A$ then $ x\in B.$
$\rightarrow:$
If $x \not\in B$ then $x \not\in A$, I.e.
$ x\in B^c$ then $x \in A^c$, or
$B^c \subset A^c$.
A: You are on the right lines, but as you said yourself - the logic isn't quite there. You want to take what you are intuitively saying and turn it into formal logic.
We want to show that $C - B \subseteq C - A$. In other word we want to show that every element of $C - B$ is an element of $C - A$.
So start by showing "If $x \in C - B$, then ..., then $x \in C - A$". Once you have done this you can conclude that $C - B \subseteq C - A$ as $x$ was an arbitrary element.

As for how you go about filling in "If $x \in C - B$, then ..., then $x \in C - A$", think about what it means for an element to be in $C - B$ and use the fact that $A \subseteq B$.
A: Element chase.
We want to show $x\in C\setminus B \implies x\in C \setminus A$.
In this context, talking it out will have everything fall into place.
If $x \in C\setminus B$ then $x \in C$ and $x \not \in B$.  If $x \not \in B$ then $x \not \in A$. (Why?)  So $x \in C$ and $x \not \in A$.  So $x\in C \setminus A$.  And $ C\setminus B \subset C \setminus A$.
There is a bit of a stickler in that we should prove that if $X \subset Y$ then $x\not \in Y \implies x\not \in X$.  The definition of $X \subset Y$ is $x\in X \implies x\in Y$.  So the contrapositive of the definition is $x \not \in Y \implies x\not \in X$.  That's enough.  If it's seems too glib, there is a formal proof by contradiction that:
If $X\subset Y$ and $x \not \in Y$.  If $x \in X$ then $x \in Y$ and that is a contradiction so $x\not \in X$.
