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?

(1) Suppose $a \in A$. Since $X = A \cup B$, we know $a \in X$, by definition of union. Moreover, since $A \cap B = \emptyset$, we know $a \in A$ implies $a \not\in B$. Thus, $a\in X$ and $a \not\in B$ implies $a \in X\backslash B$, by definition of set minus. Hence, $A \subseteq X\backslash B$.
Now, suppose $x \in X\backslash B$, i.e. $x \in X$ and $x \not\in B$. Recall $X = A \cup B$, so that $x \in X$ implies $x \in A$ or $x \in B$. Moreover $x \not\in B$, so it must be $x \in A$. Hence, $X\backslash B \subseteq A$.
It follows that $A = X \backslash B$. A similar argument shows $B = X\backslash A$.
