How do you show that two sets are disjoint? Here's a problem I am trying to solve for recreation. 
$$
A\cap B\subset C' \text{ and } A\cup C\subset B. \hspace{2 mm}\text{    Show that $A$ and $C$ are disjoint.}
$$
I can clearly see how A and C would be disjoint. Essentially, if my understanding is correct, A and C are non-overlapping sets within the bound of set B. But, I'm not exactly clear on how I would prove this by set logic. 
If you could provide some guidance, I would really appreciate it. 
 A: The second condition says that all of $A$ is inside $B$. So $A\cap B = A\subset C'$.
A: Suppose that $A \cap C \neq \emptyset $, and let $x \in A$ and $x \in C$, thus $x \in A\cup C$, and then $x \in B$, then $x \in A \cap B$, thus $x \in C'$. So $x \notin C$, contradiction $x \in C$.
A: Let's try to capture your argument algebraically.
I think your core idea is that if you take $B$ as the universe, then $A$ and $C$ are still subsets of the universe, and $A \cap C$ within $B$. The question, now, is how to translate that back to the actual universe you're working in.
The main thing you want here is that taking the intersection of everything with $B$ is how you get from the whole universe to within $B$, and this leaves unchanged the sets already within $B$.
That is, if $S \subseteq B$, then $S \cap B = S$.
So that's what you want to use. Prove first that


*

*$A = A \cap B$ (or equivalently $A \subseteq B$)

*$C = C \cap B$ (or equivalently $C \subseteq B$)


and then your statement that they are disjoint "within $B$" becomes


*

*$(A \cap B) \cap (C \cap B) = \varnothing$


If you can prove that too, then you put all three bullets together and you win.
The first two bullets are easy to prove: e.g. $A \subseteq A \cup C \subseteq B$. The last bullet is straightforward too:
$$ (A \cap B) \cap (C \cap B) \subseteq C' \cap (C \cap B) = (C' \cap C) \cap B = \varnothing \cap B $$
A: $\def\wi\{\subseteq}$$A,C \wi B$ by the second condition and hence $A \cap C = ( A \cap B ) \cap C \wi C' \cap C = \varnothing$ by the first.
But if you want to prove from first principles: Take any $x \in A \cap C$. Then $x \in A$, so $x \in A \cup C$, and hence $x \in B$. Thus $x \in A \cap B$, so $x \in C'$. But we also have $x \in C$. Therefore there is no such $x$.
