Proof: $A\subseteq (B\cup C)$ and $B\subseteq (A\cup C)$ then $(A - B) \subseteq C$ How do I prove this:
Let $A, B$ and $C$ be sets, $A \subseteq  (B \cup  C)$ and $B \subseteq  (A \cup  C)$ then $(A - B) \subseteq  C$
How about this: 
Let $x \in A$ and $y \in  B.$
Since $A \subseteq  (B \cup  C)$ then $x \in  (B \cup  C).$
Since $B \subseteq  (A \cup  C)$ then $y \in  (A \cup  C)$
$x \in  B \cup  C$, so if $x \in  B$, then $x \notin  A - B.$
if $x \notin  B$, then $x \in  C.$
$y \in  A \cup  C$, so if $y \in  A$, then $y \notin  A - B.$
if $y \notin  B$, then $y \in  C.$
So in both cases, $A - B \subseteq  C$
Is it correct? if yes, the converse is false right?
 A: The question only asks you to prove that $A-B$ is a subset of $C$. This means that we need to take $x\in A-B$ and show that $x\in C$.
Let $x$ be such element, then $x\in A$ and therefore $x\in B\cup C$. However, $x\notin B$ and therefore $x\in C$.

In the other direction, it is true that if $A-B\subseteq C$ then $A\subseteq B\cup C$. See if you can prove it.
(Hint: $x\in A$ then either $x\in B$ or $x\notin B$.)
A: You stated more than the necessary assumptions, as pointed out by a comment and an answer that only used $A \subseteq B \cup C$.  You can check the converse if you only need to prove the condition $A \subseteq B \cup C$.  So for the "full converse" the way you stated it, the only additional question that remains is whether $A-B \subseteq C$ implies $B \subseteq A \cup C$.  Hint: Consider $C$ to be the empty set, and when $A \subset B$, where $A \neq B$.
A: Well... The converse will be this:
Let A,B and C be sets, (A−B)⊆C then A⊆(B∪C) and B⊆(A∪C)
I think this is false. Take the negation and proof:
Negation: Let A,B and C be sets, (A−B)⊆C then A Not⊆ (B∪C) or B Not⊆ (A∪C) 
Here is a counter example: 
A = {1, 2}
B = {1, 4}
C = {2, 5}
So A - B = {2} and is a subset of C.
But, A U C = {1, 2, 5} - but B = {1, 4}. So B is not a subset of AUC. Element 4 is missing in AUC.
Correct?
