Assume first that $A \subseteq B$. If $A \cup B = \emptyset$, then $A \cup B = \emptyset \subseteq B$, so we may assume that $A \cup B \neq \emptyset$. If $x \in A \cup B$, then $x \in A$ or $x \in B$. If $x \in A$, then $x \in B$ since $A \subseteq B$, so that $x \in B$ in either case. Therefore, $A \cup B \subseteq B$. Conversely, result R4145: Binary union is an upper bound to both sets in the union shows that $B \subseteq A \cup B$. Therefore, $A \cup B = B$.

Conversely, assume that $A \cup B = B$. Then for every $x \in A$, we have $x \in B$, which by definition means that $A \subseteq B$. $\square$