Prove that $\displaystyle (A-B) \cap C = \emptyset \Leftrightarrow A \cap C \subseteq B $.

Proof: Suppose that $\displaystyle (A-B) \cap C \neq \emptyset $. If $\displaystyle x \in A \cap C $ then $\displaystyle x \in A $ and $\displaystyle x \in C \Rightarrow x \in B $. Then $\displaystyle \exists x \in (A-B) \cap C $ such that $\displaystyle x \in A $ and $\displaystyle x \in C $ but $\displaystyle x \not \in B $. But this is a contradiction and so $\displaystyle (A-B) \cap C = \emptyset $. How would I go about proving that $\displaystyle A \cap C \subseteq B $?

I've tried writing $\displaystyle (A-B) \cap C = \emptyset $ as $\displaystyle (A \cap B') \cap C = \emptyset $ and then $\displaystyle C \subseteq (A \cap B') $ or $\displaystyle C \cup (A \cap B') = A \cap B' $. But from here I get stuck.

Any help is appreciated.

Sorry for sounding like an idiot but I am learning the material by myself for fun and have no one to guide me.

Thanks