Basic set theory proof of implication Prove if $A \cup C \subseteq A \cap C$ then $A=C$
I'm attempting this proof and am having second thoughts that I am right.
Attempt(Contrapositive):
Assume $A \neq C$ 
Assume WLOG $x \in A$ and $x \notin C$
Then $x \in A \cup C$ but $x \notin A \cap C$ thus $A \cup C \not\subset A \cap C$
I felt like this was the correct way to prove this but I think I was supposed to assume $x \notin A \cup C \implies x \notin A \cap C$
Are both ways correct?
 A: Your first proof checks out! For the contrapositive, $A\cup C \not\subseteq A\cap C$ is the conclusion you have to make, so you do not need to make any assumptions about it. Note you can also show this directly by showing $A \subseteq C$ and $C \subseteq A$, assuming $A \cup C \subseteq A \cap C$.
Let $x \in A$. By the hypothesis, $$x \in A \cup C \subseteq A \cap C \implies x \in A \cap C \implies x \in C.$$ Since $x \in A$ was chosen arbitrarily, you have $A \subseteq C$. Similarly, for any $x \in C$, the same argument holds that $$x \in C \implies x \in A \cup C \implies x \in A \cap C \implies C \subseteq A.$$ 
A: As others have noted, your original attempt was technically correct in terms of your thoughts (which are really what matter!), but maybe we can go about piecing together a better solution using the basic algebra of sets. 
First, we can prove a stronger statement than what you were originally required to prove, namely we can prove an "if and only if" version of your statement where the direction currently not under consideration simply reduces down to showing that a set is a subset of itself. Great. 
Second, let's try to prove such an if and only if statement using only basic set algebra:
\begin{align}
A\cup C\subseteq A\cap C
&\iff x\in A\cup C\to x\in A\cap C\\[0.5em]
&\iff x\not\in A\cup C\lor x\in A\cap C\\[0.5em]
&\iff (x\not\in A\land x\not\in C)\lor(x\in A\land x\in C)\\[0.5em]
&\iff [(x\not\in A\land x\not\in C)\lor x\in A]\land[(x\not\in A\land x\not\in C)\lor x\in C]\\[0.5em]
&\iff [(x\not\in A\lor x\in A)\land(x\not\in C\lor x\in A)]\land[(x\not\in A\lor x\in C)\land(x\not\in C\lor x\in C)]\\[0.5em]
&\iff (x\not\in C\lor x\in A)\land(x\not\in A\lor x\in C)\\[0.5em]
&\iff (x\in C\to x\in A)\land(x\in A\to x\in C)\\[0.5em]
&\iff C\subseteq A\land A\subseteq C\\[0.5em]
&\iff A=C.
\end{align}
It may not be the prettiest in some ways, but hopefully it illuminates another way of going about the problem.
