Is this set statement true? $$B \cap C \subseteq A \implies (C-A) \cap (B-A) = \varnothing.$$
I don't think this is true because B and neither C are necessarily a subset of A. Only B intercept C is a subset of A.
 A: $$B \cup C \subset A \implies B \subset A \text{ and } C \subset A$$ Hence, $$C - A = \emptyset = B-A$$
EDIT
Since the OP changed the question to $B \cap C \subset A$, instead of $B \cup C \subset A$, below is the revised answer.
$$(C-A) \cap (B-A) = (C \cap A^c) \cap (B \cap A^c) = (B \cap C) \cap A^c \subseteq A \cap A^c = \emptyset$$
A: Revised to match the corrected problem statement:
If $B\cap C\subseteq A$, then $(B\cap C)\setminus A=\varnothing$. But $$(B\cap C)\setminus A=(B\setminus A)\cap(C\setminus A)\;,\tag{1}$$ 
so if $(B\cap C)\setminus A=\varnothing$, then $(B\setminus A)\cap(C\setminus A)=\varnothing$.
Proving $(1)$ is a good exercise, and not hard.
A: The statement is true. 
We prove the contrapositive. 
Suppose that $(B\setminus A)\cap (C\setminus A) \neq \emptyset$. Then there is some $x \in (B\setminus A) \cap (C\setminus A)$. By definition $x\notin A$, however 
$(B\setminus A) \cap (C\setminus A) \subset B \cap C$. So we have $x \in B\cap C$ and $x \notin A$, 
So $B\cap C$ is not a subset of $A$. 
EDIT: To expand this explaining every step more carefully, hopefully this helps. 
Proving the contrapositive means we show that $\neg (C\setminus A) \cap (B\setminus A) = \emptyset \rightarrow \neg B\cap C \subset A$, which is equivalent to the original statement. 
If we assume $(B\setminus A) \cap (C\setminus A) \neq \emptyset$ then this means that this set has an element, so we take some $x\in (B\setminus A) \cap (C\setminus A)$. Then since $(B\setminus A)\cap (C\setminus A)$ is the set of elements which are in $B$ and $C$ and not in $A$, this means that $x\in B\cap C$ and $x\notin A$. Therefore we have produced an element, $x$, which is in $B\cap C \setminus A$ which means that $B\cap C$ can't be a subset of $A$.
A: With the intersection, think about what the equation means. Suppose you have an element of B which is not in A, can it be in $B\cap C$? Can it be in $C$? Can it be in $C-A$?
A: $(C \setminus A) \cap (B \setminus A) = \emptyset$ means that all elements are in   $((C \setminus A) \cap (B \setminus A) )'=(C-A)'\cup(B-A)'$ by De Morgan,
$=(C'\cup A) \cup (B' \cup A)$ combining an alternate definition of $A \setminus B$ and De Morgan again,
$=(C'\cup B')\cup(A\cup A)=(C\cap B)' \cup A$
Since $B \cap C \subseteq A$, $A' \subseteq (B\cap C)'$, so our union is indeed the whole set. 
