$A\subseteq B\to C\setminus B\subseteq C\setminus A\,$ -- how to prove this? Given $A \subseteq B $. Prove for every set $C, C\setminus B \subseteq C \setminus A  $.
Logical Argument:
Given: $\forall x,  x \in A  \rightarrow x \in B $
Goal: $\forall C \forall x , x\in C\setminus B \rightarrow x \in C \setminus A$
For arbitrary $C$ and $x$,
  $  x\in C\setminus B \rightarrow x \in C \setminus A$
rewriting,
    $  x\in C\ \wedge x \not \in B \rightarrow x \in C  \wedge \not \in A$
adding,
$  x\in C\ \wedge x \not \in B $ to the given.
Proof by contradiction:
Now the given contains,
$ x \in A  \rightarrow x \in B $
$  x\in C\ \wedge x \not \in B $
$x  \not \in C  \vee x \in A$
In the disjunction, if I choose $x \not \in C $, then it's a contradiction.  If I choose $x \in A $, then agin it's a contradiction because modus ponens, it adds $x \in B$ which is again a contradction.
Is there any better way to argue this?
 A: 
Is there any better way to argue this? 

I would argue it directly.
Claim: $A\subseteq B\to C\setminus B\subseteq C\setminus A$ for all sets $C$. 
Proof. Suppose $A\subseteq B$. Further suppose $x\in C\setminus B$, where $x$ is arbitrary. If $C=\varnothing$, then $x\in C\setminus A$ is trivially true (i.e., $C\setminus B\subseteq C\setminus A$ is vacuously true when $C=\varnothing$). Suppose, then, that $C\neq\varnothing$ and $x\in C\setminus B$. By definition, $x\in C\setminus B$ means $x\in C\land x\not\in B$. Since we are given $A\subseteq B$, we observe that $x\in A\to x\in B$ is logically equivalent to $x\not\in B\to x\not\in A$ (contrapositive). Thus $x\in C\land x\not\in B$ implies that $x\in C\land x\not\in A$, and the definition of $C\setminus A$ is $x\in C\land x\not\in A$. Thus, $x\in C\setminus A$. Hence, if $A\subseteq B$, then $C\setminus B\subseteq C\setminus A$. $\blacksquare$
A: $$
A \subseteq B \Rightarrow B^c \subseteq A^c \Rightarrow C\cap B^c \subseteq C \cap A^c \cong C\setminus B \subseteq C \setminus A
$$
