For all sets $A$, $B$, and $C$, if $A-B \subseteq A - C$ then $ A \cap C = \varnothing $ Prove the statement P: 

For all sets $A$, $B$, and $C$, if $A-B \subseteq A - C$ then $ A \cap C = \varnothing $

My attempt to answer:
This statement is true, and here is a proof:
Proof: Suppose A, B, and C are sets such that $A-B \subseteq A - C$. We want to prove that $A \cap C = \varnothing $ by contradiction.
Suppose $A \cap C \neq \varnothing $. That is there is $x \in A \cap C$. Also, since $A-B \subseteq A - C$, suppose there is a $x \in A - B$, then $x \in A - C$. That is, $x \notin C$. Therefore, that implies that $x \notin A \cap C$ which contradicts with given that $x \in A \cap C$. Therefore, concluding by contradiction: $A \cap C = \varnothing $. End of proof.
Is this proof correct? Any issues?.
My proof is wrong, got it based on your comments:
Let $A = \{1, 2\}, B = C = \{2\}$, then $A-B \subseteq A - C$, but, $A \cap C \neq \varnothing = \{2\}$. Can someone, tell me what I did wrong in my proof above? Is it because, I assumed $x \in A - B$?
part b) Write the converse and proof/disproof that. 
The converse is: For all sets $A$, $B$, and $C$, if $ A \cap C = \varnothing $ then $A-B \subseteq A - C$
 A: A counterexample: Let $A=B=C=\{17\}$. 
The converse is true: If $A\cap C=\emptyset$, then $A\setminus B\subseteq A\setminus C$. For if $A\cap C=\emptyset$ then $A\setminus C=A$.
Remark: The added example that tries to show the converse does not hold is not correct. For you let $A=\{1\}$ and $C=\{1\}$, which makes $A\cap C$ non-empty.
A: Before you go about trying to prove a statement that you're not absolutely sure is true, it's a good exercise to spend a bit of time trying to disprove it, that is, to try to find examples where the statement fails.
This will not only save you from a lot time spent banging your head against the desk, but it will also help you develop a sense and a set of skills to help you determine what's likely to be true and what isn't. Also, more often than not, multiple unsuccessful attempts at a counterexample can give you some ideas that can lead to a proof.
For your particular example, start of with something simple, like various sets of integers (any set of symbols will do, but integers are nice). See if you can come up with three sets of integers $A$, $B$, and $C$ such that "every integer that is in $A$ but not in $B$ is also not in $C$", but "there is at least one integer that is in both $A$ and $C$".
Here's an example (spoiler alert):

 $A = \{1,2,3\}$, $B = \{2,3\}$, $C = \{3\}$

A: What is wrong with your proof: Suppose $x \in A \cap C$ is an arbitrary element. Since $A - B \subseteq A - C$, for an arbitrary (not necessarily equal to $x$) $y \in A - B$, $y \in A - C$.
I claim here that $y$ cannot be equal to $x$. 
If, by contradiction $y = x$, then $y \in A - B$ and $y \in A-C$ and $y \in A \cap C$. Here we have by definition:
$$y \in A - B \implies y \in \{z: z \in A, z \notin B\}$$
$$y \in A - C \implies y \in \{z: z \in A, z \notin C\}$$
so $$y \in \{z: z \in A, z \notin B, z \notin C\}$$
and
$$A \cap C = \{z: z \in A, z \in B\}$$
and of course $\{z: z \in A, z \notin B, z \notin C\} \cap \{z: z \in A, z \in B\} = \varnothing$.
A: Your statement is NOT true. If you consider the sets
$$
A=\{a_1,a_2,a_3\},\quad B=\{a_1,a_2\},\quad C=\{a_1\},
$$
you can see that
$$
A-B=\{a_3\}\subset A-C=\{a_2,a_3\},
$$
but $A\cap C=\{a_1\} \ne \varnothing$
A: The question is equivalent to: use $(A-AB)(A-AC)=(A-AB)$ to prove $AC=\emptyset$(Boole Algebra)
$A(E-B)(E-C)=A-AC-AB=A-AB\Leftrightarrow A-AC+AB=A \Leftrightarrow AC\subset AB$
so we need to add additional condition $A\cap B=\emptyset$ to guarantee $A\cap C=\emptyset$
