proof that $A − B ⊆ C$ if and only if $A ⊆ B ∪ C$ proof that $A − B ⊆ C$ if and only if $A ⊆ B ∪ C$
$A − B ⊆ C$ implies $A − B ⊆ C$
suppose $A − B ⊆ C $ , let $ x \in A $ and suppose $ x \not \in B.$
so $x \in A − B$ since  $A − B ⊆ C$, $x \in C$
since $ x \in A $ and $ x \in C $ implies $A ⊆ B ∪ C$
is this proof correct?
$A ⊆ B ∪ C$ implies $A − B ⊆ C$ i will proof by cntradiction
 
suppose $A ⊆ B ∪ C$ , and let $x \in A-B$ so $ x \in A $ and $ x \not \in B$
suppose $ x \not \in C $, since $A ⊆ B ∪ C$  so $ x \in B ∪ C $
since $ x \not \in B$ ,it must be  $ x \in C $
but this contradict to assumption that $ x \not \in C $
hence $x \in A-B \in C$ 
 A: No. You've shown that $A - B \subseteq C \implies A \subseteq B \cup C$. However, you have to prove the other direction too as this is an "if and only if" statement.
Your proof for the forward direction is also a little more roundabout than it needs to be - it's correct but it feels a little clunky. For example, I would prove it as so:


*

*Suppose $A - B \subseteq C$ and consider $x \in A - B$

*By definition of set difference, $x \in A, x \not \in B$

*By our assumption, $x \in C$ as it is in a subset of $C$

*Since $x \in C,$ then $x \in B \cup C$.

*Since $x \in A$ and $x \in B \cup C$, our premise implies $A \subseteq B \cup C$.


What remains to be shown is the converse, $A \subseteq B \cup C \implies A - B \subseteq C $.

Edit:
Regarding your proof of the converse, it seems mostly correct, though I feel a proof by contradiction isn't necessary here. But I suppose if you find that easier to handle, more power to you. It would help for you to elaborate a bit more on what you're trying to do with the proof, though.
I would write up this proof as follows.


*

*Suppose $A \subseteq B \cup C$ but $A-B \not \subseteq C$.

*By the latter, take some $x \in A-B$ such that $x \not \in C$.

*Then $x \in A, x \not \in B, x \not \in C$.

*Since $x \not \in B, x \not \in C$, then $x \not \in B \cup C$.

*However, $x \in A$, and if $A \subseteq B \cup C$ then $x \in B \cup C$.

*Thus, our assumption $A-B \not \subseteq C$ is false$^{(1)}$, and thus $A-B \subseteq C$
$(1)$: I feel like there's some exception to all this regarding empty sets that needs to be accounted for, but I can't pin down what at the moment. The reason I feel this is that we can define the empty set by $\emptyset = \{x|x\in S, x\not \in S\}$ for a set $S$. Thus, perhaps we need to account for the case where $B \cup C = \emptyset$ which would mean $A = B = C = \emptyset$ - but then $A - B = \emptyset \subseteq \emptyset = C$, so I don't know if that's a problem. I'm not 100% sure on that front, and maybe I'm overthinking it. 
I do buy your proof for nonempty sets at least, and I feel like that the only case it doesn't account for is all empty sets like noted above which gets the results you desire immediately.
A: Your proof is correct, but it is only the " $\implies $ " part. You have to show the converse. That is if $ A \subseteq B \cup C$, you have to show $ A \setminus B \subseteq C$. So if $x \in A\setminus B$, then $x \in A $ and $x \notin B$. Thus $x \in B \cup C$ and $x \notin B$. So $x \in C$. Thus $A\setminus B \subseteq C$.
A: Left to do :
$A - B \subset C$ $\Rightarrow$ $A \subset B \cup C$.
Let $x \in A$:
Then $x \in B$ or $x \not \in B$.
If $x \not \in B$ , $x \in C$, since $A-B \subset C$.
Hence
$x \in B$ or $x \in C$ , i.e $x \in B \cup C$.
A: It might be interesting to see that it is not necessary  to prove the equivalence in two times ( first from left to right,then  from right to left), for here one can reach the goal susbtituting progessively equivalent proposition  for equivalent proposition.  
I add, to a first proof using logic, a second proof using only the laws and definitions of the algebra of sets. 


