YES, YES, YES! That is correct.
NO, NO, NO This not correct.
One direction: Given $\displaystyle \left( {A \cap C} \right) \subseteq B$ prove that $\displaystyle \left( {A - B} \right) \cap C = \emptyset $.
The other direction: Given $\displaystyle \left( {A - B} \right) \cap C = \emptyset $ prove that $\displaystyle \left( {A \cap C} \right) \subseteq B$.
Why is it not correct?
We assume that $\displaystyle (A-B) \cap C \neq \emptyset $ for proof by contradiction.
One direction: Given that $\displaystyle (A \cap C) \subseteq B $ and $\displaystyle (A-B) \cap C \neq \emptyset $ use proof by contradiction to show that $\displaystyle (A-B) \cap C = \emptyset $ (my thinking).
Your other direction looks good.
So now is this correct?
I asked help for the $\displaystyle \Longrightarrow $ direction.
I agree with this method that Plato has done.
For the $\displaystyle \Longleftarrow $ direction I already proved it by contradiction.
So is this not correct?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 $.
Here it is.
Given $\displaystyle \left( {A \cap C} \right) \subseteq B$ that means $\displaystyle \left( {A \cap C} \right) \cap B' = \emptyset $.
But $\displaystyle \left( {A \cap C} \right) \cap B' = \left( {A \cap B'} \right) \cap C = \left( {A - B} \right) \cap C = \emptyset$