Is a contradiction enough to prove a set equality to $\varnothing$? An exercise asks me to prove the following:
$$A \cap (B-A) = \varnothing$$
This is what I did:
I need to prove that $A \cap (B-A) \subseteq \varnothing$ and $\varnothing \subseteq A \cap (B-A)$
The second one seems to be obvious by definition (not sure if it is fine to say that in a test).
But the first one goes like this:
$$x \in A \cap (B-A)$$
$$x \in A \land x \in (B-A)$$
$$x \in A \land (x \in B \land x \notin A)$$
$$(x \in A \land x \notin A) \land x \in B$$
$$(x \in A \land x \notin A)$$
Since there is a contradiction... err... the proof... is good.
Okay, that's my problem. I think that the procedure is fine, but clearly I am unable to word it out (well, I don't even know if it is valid at all). Is it valid? How can I... "prove that my proof is correct"? I don't really see this contradiction as an obvious indicator of $A \cap (B-A) \subseteq \varnothing$

Yes this is perfectly valid. 
You derive a contradiction from the assumption that there is an $x$ satisfying $$x \in A \cap (B\setminus A),$$ so there must be no $x$ satisfying this inclusion and the set is empty. 
