Let $A,B,X$ be sets such that $A\cup B = X$ and $A \cap B = ∅$. Show that $A = X\backslash B$ and $B = X\backslash A$. I'm trying to prove this 

Let $A,B,X$ be sets such that $A\cup B = X$ and $A \cap B = ∅$. Show
  that:
  (1) $A = X\backslash B$
  and
  (2) $B = X\backslash A$.

My proof is  
Let $x \in A$. We know that $x \notin B$ by definition of intersection. It follows that $x \in A \cup B$ by definition of union so we have $x \in X$. Therefore all elements of A must not be in B and must also be in X so $A=X \backslash B$ by definition of difference sets. We can repeat the same process on an arbitrary element of B to get $B= X \backslash A$.
Is this correct and is there any way I could word my argument better?

This argument is a correct but is not complete. In your proof of (1) you proved only that if $x \in A$ then $x \in X\backslash B$. This is just half of proving (1).    
To have a complete proof of (1) you need to also prove the reverse implication i.e. you need to prove that if $x \in X\backslash B$ then $x \in A$.  
The statement (2) you don't really need to prove, it follows by symmetry as you noticed.  
