Proving or disproving if sets are equal 
Does this mean we can ignore A because its a subset of B, so now were only dealing with B or C? So you substitute B or C with A. Am I right because that gives me the right answer. 
(C-C) and empty set = empty set
or 
(B-b) and empty set = empty set.
Please help, I made a reasonable attempt on this problem
 A: Okay you seem to have a very feeble grasp on the fundamentals of Set Theory so I suggest you first try and read these 12 pages before reading my answer. 
I will use the following rules and definitions in proving the answer. 


*

*$A \subseteq B \iff (x \in A \implies x \in B)$

*$A \cup B = \text {the set of all $x$ such that $x \in A$ or $x \in B$ } $

*$A \cap B = \text {the set of all $x$ such that $x \in A$ and $x \in
   B$ } $

*$A - B = \text {the set of all $x$ such that $x \in A$ but $x \not \in B$}$


Let there be an element $x$ in $(A-C) \cap(A-B)$. 
Then $x \in (A-C) $ and $x \in (A - B)$. Then we have the two statements - "$x$ is in $A$ but not in $C$" and "$x$ is in $A$ but not in $B$" - both of which must be satisfied. Assume the first statement is true. That is $x$ is in $A$ but not in $C$. The only way the second one too is satisfied is if $x$ is not in $C$. That is $x$ must be in $A$ but not in $B$ and not in $C$ i.e. $x$ must be in $A$ but not in $B$ or $C$. Can this occur? NO!! Why?
Because we are given that $A \subseteq B \cup C$. This is true if and only if the statement "If $x$ is in $A$ then $x$ is in $B \cup C$" which contradicts our development in the previous paragraph. Thus we arrive at a contradiction that says our initial assumption was false. What was that?
That there was an element $x$ in $(A-C) \cap(A-B)$. 
Thus we assert that $(A-C) \cap(A-B)$ is empty. 
Q.E.D.
A: I think you can do it symbolically by just "subtracting" both sides of the hypothesis inclusion from $A$ and reversing the inclusion:
$$A\subset B\cup C \implies A-(B\cup C) \subset A-(A)=\varnothing$$
Then use DeMorgan's Law to write
$$(A-B)\cap(A-C)\subset\varnothing$$
which clearly implies
$$(A-B)\cap(A-C)=\varnothing$$
