Set Theory Contradiction Proof Verification Let A, B and C be subsets of a universal set U.
Prove by contradiction that $$A\cap B \subseteq C \to (A-C)\cap B = \emptyset$$
Suppose otherwise, $A\cap B\subseteq C \land (A-C)\cap B \neq \emptyset$. 
Let $n\in A$, then by definition of the subset, $n\in C$. Since $n\in A \land n\in C, n\not \in (A-C)$, by definition of the set difference, this means $(A-C)= \emptyset$. Therefor, $n\not \in (A-C)\cap B$, by definition of the intersection. Thus, by definition of the empty set, $(A-C)\cap B = \emptyset$. This is a contradiction. Thus, $A\cap B \subseteq C \to (A-C)\cap B = \emptyset$ must be true.
Can someone tell me whether I did this right. I think it makes sense, but it also seems like I made a mistake somewhere because it seems too easy and short. Thank you.
 A: You have proven that, for every $n\in A$, $n\notin (A-C)\cap B$. That does not make $(A-C)\cap B$ the empty set, since you haven't proven that it has no elements, only that it has no elements that belong to $A$. 
The answer is even shorter than yours: suppose that $(A-C)\cap B\neq\emptyset$. Then, there exists some $x\in (A-C)\cap B$. This means that $x\in A$, $x\notin C$ and $x\in B$, that is, $x\in A\cap B$ but $x\notin C$. This proves by definition of inclusion that the statement $A\cap B\subseteq C$ is false. Since we find that $(A-C)\cap B\neq\emptyset\Rightarrow A\cap B\not\subseteq C$, we conclude that $(A\cap B)\subseteq C\Rightarrow (A-C) \cap B=\emptyset$. 
A: We prove following:

(a)  $(A \cap B)-C=(A-C)\cap B$
$$\begin{align} x \in (A \cap B ) - C &\leftrightarrow x \in (A \cap B) \wedge x \notin C \\ &\leftrightarrow (x \in A \wedge x \in B)\wedge x \notin C \\ &\leftrightarrow x \in A \wedge (x \in B\wedge x \notin C )\\ &\leftrightarrow x \in A \wedge (x \notin C \wedge x \in B) \\ &\leftrightarrow (x \in A \wedge x \notin C) \wedge x \in B \\ &\leftrightarrow x \in (A -C) \wedge x \in B \\ &\leftrightarrow x \in (A -C)\cap B \end{align}$$
(b) $(A \cup B)\setminus B=A \setminus B$
  \begin{align*} x \in(A \cup B)\setminus B &\leftrightarrow x \in (A\cup B) \wedge x \notin B \\ &\leftrightarrow (x \in A \vee x \in B)\wedge x \notin B\\ &\leftrightarrow (x \in A \wedge x \notin B) \vee (x \in B \wedge x \notin B) \\ &\leftrightarrow x \in (A\setminus B) \vee x \in (B\setminus B) \\  &\leftrightarrow x \in ((A\setminus B) \cup \emptyset )\\ &\leftrightarrow x \in (A\setminus B) \end{align*}

We prove

(c) $A\cap B \subseteq C \to (A-C)\cap B = \emptyset$
direct 
  $$ \begin{align} (A-C)\cap B &= (A \cap B)-C \text{ (by (a))}  \\ &=((A\cap B) \cup C) -C \text{ (by (b))}\\ &= C -C \text{ (because } A\cap B \subseteq C) \\&= \emptyset \end{align} $$
by contrapositive 
  $$\begin{align} (A -C)\cap B \neq \emptyset &\to \exists x : x\in (A -C)\cap B \\&\to \exists x :x \in (A \cap B) -C \text{ (by (a))}\\ &\to \exists x : (x \in A \wedge x \in B )\wedge x \notin C \text{ (by Definition)}\\ &\to A\cap B \nsubseteq C \end{align}$$
by contradiction 
  $$\begin{align} (A -C)\cap B \neq \emptyset &\to \exists x : x\in (A -C)\cap B \\&\to \exists x :x \in (A \cap B) -C \text{ (by (a))}\\ &\to \exists x : x \in C \wedge x \notin C \text{ (by hypothesis)}\\ &\to \exists x: x \in \emptyset \text{ (Absurd!)} \end{align}$$

