Given below is an attempted proof of a result.
Proof First, we show that $A \subseteq(A \cup B)-B$. Let $x \in A$. Since $A \cap B=\emptyset$, it follows that $x \notin B$. Therefore, $x \in A \cup B$ and $x \notin B$; so $x \in(A \cup B)-B$. Thus $A \subseteq(A \cup B)-B$.
Next, we show that $(A \cup B)-B \subseteq A$. Let $x \in(A \cup B)-B$. Then $x \in A \cup B$ and $x \notin B$. From this, it follows that $x \in A$. Hence $(A \cup B)-B \subseteq A$.
(a) What result is being proved above?
(b) What change (or changes) in this proof would make it better (from your point of view)?