$A ⊂ B$ if and only if $A − B = ∅$ I need to prove that $A ⊂ B$ if and only if $A − B = ∅$.
I have the following "proof":
$$ A \subset B \iff A - B = \emptyset$$
proof for $\implies:$
$$\forall x \in A, x \in B$$
Therefore,
$$A - B = \emptyset$$
proof for $\impliedby$:
If $$A - B = \emptyset$$
then
$$\forall x \in B, x \in A$$
since $\forall x \in B, x \in A$, 
$$ A \subset B $$
However the whole thing seems to be incredibly "fragile" and relies on circular logic (see how I just switched the sets in the for all statements)
Is this a valid proof? Is there a better way to write it?

I think your proof is Ok. I just add more details.
$ (\Rightarrow) $; Suppose $ A\subset B $. Then $ \forall x\in A,x\in B $. Therefore if $ x\notin B $ then $ x\notin A $. That is if $ x\in B^{c} $ then $ x\in A^{c} $. Since $ A-B=A\cap B^{c} $ we have that $ A-B=\varnothing $.
$ (\Leftarrow) $; Conversely suppose $ A-B=\varnothing $. So $ A\cap B^{c}=\varnothing $. Therefore $ \forall x\in A,x\notin B^{c} $. Hence $ \forall x\in A,x\in B $. So we have that $ A\subset B $.
