Possible implications of $ ( A \cap B ) \subset ( C \cap D ) $ I was wondering; if 
$$ ( A \cap B ) \subset ( C \cap D ) $$ 
Can anything be said about the relationship between $( A \cap B )$ and:
$$( A \cap C )$$
$$( A \cap D )$$
$$( B \cap C )$$
$$( B \cap D )$$
 A: 
Edit: I received some downvotes and "this doesn't answer the question" comments which I don't really understand unless I grossly misinterpreted the question.

Here is my interpretation. @OP, please let me know if this is not what you intended.

If we assume that $(A \cap B) \subset (C \cap D)$, then can anything be said about the relation between $(A \cap B)$ and any of $(A \cap C)$, $(A \cap D)$, $(B \cap C)$, and $(B \cap D)$?

Assuming this rephrasing of the question is valid, here is my answer.

We have:
$$A \cap B \subset A$$
$$A \cap B \subset B$$
and
$$A \cap B \subset C \cap D \subset C$$
$$A \cap B \subset C \cap D \subset D$$
Therefore, $A\cap B$ is contained in each of the sets $A$, $B$, $C$, and $D$, so it is contained in any intersection of these sets.

P.S. In response to the comment by G Cab, I interpret $\subset$ to mean "is a (not necessarily proper) subset of".
A: It depends exactly what is meant by "$\subset$." I've seen that symbol used to denote both proper subset, and possibly improper subset.
First interpretation ($\subsetneq$): Suppose $A=B$ and $C=D$, and $A\subsetneq C$. Then the hypothesis holds, but we do not have $A\cap B\subsetneq A\cap C$ (and similarly for the other three pieces of the question).
Second interpretation ($\subseteq$): Since $A\cap B\subseteq C\cap D$, we have $A\cap B\subseteq C$. Thus $A\cap (A\cap B)\subseteq A\cap C$, that is, $A\cap B\subseteq A\cap C$. The other three pieces of the question are similar.

That is: from the (a priori stronger) hypothesis $A\cap B\subsetneq C\cap D$, all we can conclude is that $A\cap B\subseteq A\cap C$ (and etc.). The latter inclusion may be improper.
A: Unfortunately, the original answer by @Bungo was correct but for some reason he was pressured to delete it.  If he decides to answer again, I encourage people to upvote that answer and for the question asker to select it as the accepted answer.

As a foreword, one of the comments on @Bungo's post was about the distinction between the use of the symbols $\subset$ and $\subseteq$.  In some contexts $X\subset Y$ implies that $X\subseteq Y$ and $X\neq Y$, i.e. that $X$ is a proper subset of $Y$.  However, in many other contexts we do not make a distinction between the two.  That is to say, depending on the author, $X\subset Y$ does not imply $X\neq Y$.  Notably, Walter Rudin's Principles of Mathematical Analysis follows the convention that $\subset$ does not imply being a proper subset.


Definition 1: A set $X$ is said to be a subset of the set $Y$, written with symbols as $X\subseteq Y$ or as $Y\supseteq X$, if and only if every element which happens to be an element of $X$ also happens to be an element of $Y$.

To prove that a set $X$ is a subset of a set $Y$, one can take a generic element of $X$, often denoted as $x$, and show that certain properties of the sets in question imply that it is also an element of $Y$.  If we did not specify anything about the element $x$ apart from it being an element of $X$, this would imply that it is true for all elements in $X$, which implies the result.

Definition 2: The intersection of two sets $X$ and $Y$, written as $X\cap Y$, is the set of all elements which are in both $X$ and $Y$ simultaneously.  That is to say, $x\in X\cap Y$ if and only if $x\in X$ and $x\in Y$.


Given that $A\cap B\subseteq C\cap D$, let $x\in A\cap B$.
Then by definition 2, $x\in A$ and $x\in B$.
Furthermore, since $x\in A\cap B$ and $A\cap B\subseteq C\cap D$ by definition 1 this implies that $x\in C\cap D$.  Then by definition 2, this further implies that $x\in C$ and $x\in D$.
Thus, for any $x\in A\cap B$ it follows that all of the following are true: $x\in A$ and $x\in B$ and $x\in C$ and $x\in D$.  Thus, $x\in A\cap B\cap C\cap D$ and therefore by definition 1 $A\cap B\subseteq A\cap B\cap C\cap D$.
In particular then, by looking at only two of the four statements at a time, this implies that $x\in A\cap B, x\in A\cap C, x\in A\cap D, x\in B\cap C, x\in B\cap D$ and $x\in C\cap D$.
Then by definition 1, one has $A\cap B$ is a subset of all of the sets listed in question.
