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$B \cap C \subseteq A \implies (C-A) \cap (B-A) = \varnothing.$

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.

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    Just so you're clear and for future reference: $B \cap C$ reads: "B intersect C"; i.e., the word to use is "intersect", not "intercept"...Also, to clear up any confusion: $(C - A)$ is often/usually denoted: "$(C \setminus A$)". (In Tex, that's (C \setminus A), enclosed in "$" signs here at MathSE.)2012-10-30

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$B \cup C \subset A \implies B \subset A \text{ and } C \subset A$ 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. $(C-A) \cap (B-A) = (C \cap A^c) \cap (B \cap A^c) = (B \cap C) \cap A^c \subseteq A \cap A^c = \emptyset$

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    @DanielLittlewood THE OP had $B \cup C$ initially instead of $B \cap C$. Will change it now.2012-10-30
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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.

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    @Mathilda: You know that if $X\subseteq Y$, then $Y\setminus X=\varnothing$, right? I’m using that and $(1)$, which you can prove by element-chasing: if $x\in(B\cap C)\setminus A$, then $x\in B\cap C$ and $x\notin A$. Since $x\in B$ and $x\notin A$, we have $x\in B\setminus A$, and since $x\in C$ and $x\noting A$, we have $x\in C\setminus A$. Combining the two, we have $x\in(B\setminus A)\cap(C\setminus A)$. It’s just as easy to verify that if $x\in(B\setminus A)\cap(C\setminus A)$, then $x\in(B\cap C)\setminus A$, thereby verifying $(1)$.2012-10-30
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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 $A$.

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    @MathildaPitt I have edited hopefully it is more clear.2012-10-30
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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 $C-A$?

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$(C \setminus A) \cap (B \setminus A) = \emptyset$ 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.