Set theory proof; (subsets, universal set) 
Assume $A,B,C$ are subsets of the universal set, $U$. Prove that $$(A\cap B)\cup C=A\cap(B\cup C)$$ iff $C$ is a subset of $A$. 

How would you go about proving this? Can I do something with De Morgans law and distribute the terms? Or should I do this by choosing an arbitrary element $x$ that exists in the subsets?
 A: Just for fun, here is another way of proving this: expand the definitions, then use the laws of logic to simplify.
$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\Tag}[1]{\text{(#1)}}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$We calculate for any sets $\;A,B,C\;$:
$$\calc
(A \cap B) \cup C \;=\; A \cap (B \cup C)
\calcop\equiv{set extensionality; definitions of $\;\cap,\cup\;$, twice}
\tag{*} \langle \forall x :: (x \in A \land x \in B) \lor x \in C \;\equiv\; x \in A \land (x \in B \lor x \in C) \rangle
\calcop\equiv{logic: distribute $\;\land\;$ over $\;\lor\;$ in right hand side}
\langle \forall x :: (x \in A \land x \in B) \lor x \in C \;\equiv\; (x \in A \land x \in B) \lor (x \in A \land x \in C) \rangle
\calcop\equiv{logic: $\;\lor\;$ distributes over $\;\equiv\;$}
\langle \forall x :: (x \in A \land x \in B) \lor (x \in C \;\equiv\; x \in A \land x \in C) \rangle
\calcop\equiv{logic: simplify $\;P \equiv P \land Q\;$ to $\;\lnot P \lor Q\;$}
\langle \forall x :: (x \in A \land x \in B) \lor x \not\in C \lor x \in A \rangle
\calcop\equiv{logic: use negation of $\;x \in A\;$ on other side of $\;\lor\;$}
\langle \forall x :: (\text{false} \land x \in B) \lor x \not\in C \lor x \in A \rangle
\calcop\equiv{logic: simplify}
\langle \forall x :: x \not\in C \lor x \in A \rangle
\calcop\equiv{logic: write $\;\lnot P \lor Q\;$ as $\;P \Rightarrow Q\;$; definition of $\;\subseteq\;$}
C \subseteq A
\endcalc$$
Note that we did not need to prove two separate directions, and instead used 'iff' ($\;\equiv\;$) throughout this calculation.
Also note that at point $\Tag{*}$ we could also have chosen to distribute $\;\lor\;$ over $\;\land\;$ in the left hand side.  That would have changed the details of the calculation, but the result would be the same.  All other steps are really forced by our desire to simplify.
A: $(A \cap B) \cup C = A \cap (B \cup C) \Longleftrightarrow C \subseteq A$
$(\Rightarrow)$  $x \in C \Rightarrow  x\in (A \cap B) \cup C \Rightarrow x \in A \cap (B \cup C) \Rightarrow x \in A$
$\therefore C \subset A$
$(\Leftarrow)$ $ x \in A \cap (B \cup C) \Leftrightarrow x \in (A \cap B)\cup(A \cap C) \Leftrightarrow x \in (A \cap B) \cup C$
This last step uses $C \subseteq A \Leftrightarrow A \cap C = C$
A: The "only if" results from $(A \cap B) \cup C \subseteq A \cap (B \cup C )$ since $$C \subseteq ( A \cap B) \cup C \subseteq A \cap (B \cup C ) \subseteq A.$$
Aside: It seems that if this happens for any one $B$ then it happens for all $B.$ I feel there is more that than I see.
For the "if" part: Assume that $C \subseteq A.$ Since you seem content to use DeMorgans laws, we can then say $$ (A \cap B) \cup C =  (A \cup C) \cap (B \cup C)=A \cap (B \cup C).$$ That does it. 
We could equally well say $$A \cap (B \cup C)= (A \cap B) \cup (A \cap C) = (A \cap B) \cup C.$$
