How to prove $A \subset B \iff A \cup B = B$ Let A, B be two sets. Prove that $A \subset B \iff A \cup B = B$

I'm thinking of using disjunctive syllogism by showing that $\neg \forall Y(Y \in A).$ However, I'm not sure how the proving steps should proceed such that it leads me to that premise.
Edit: Thanks for the input. FYI, I need to prove this using predicate logic.
Taking the "longer" road. Let us review the definitions:


$A\subseteq B$ if and only if for every $x\in A$, $x\in B$.
$x\in A\cup B$ if and only if $x\in A$ or $x\in B$.
$A=B$ if and only if $A\subseteq B$ and $B\subseteq A$.
$P\iff Q$ means that if we assume that $P$ holds, then $Q$ must hold; and vice versa.




Now assume $A\subseteq B$. This means that for every $x\in A$ we have $x\in B$. We want to show that $A\cup B=B$. 


So we take $x\in B$, then $x\in A$ or $x\in B$, and therefore $x\in A\cup B$.
Now take $x\in A\cup B$, we want to show that $x\in B$. By definition either $x\in A$ or $x\in B$.


If $x\in B$ we are done.
If $x\in A$ then by the assumption that $A\subseteq B$ we have that $x\in B$.


Either way we have that $x\in B$.


We have shown that if $A\subseteq B$ then $A\cup B\subseteq B$ and $B\subseteq A\cup B$, which is by fact number $3$ to say $A\cup B=B$.



Now we need to assume that $A\cup B=B$, and deduce that $A\subseteq B$. So we need to show that if $x\in A$ then $x\in B$.

Take $x\in A$ to be an arbitrary element. Because $x\in A$ we have that $x\in A$ or $x\in B$, and therefore $x\in A\cup B$. The assumption was, however, that $A\cup B=B$ and therefore we have that $x\in B$ as wanted.
Let  $A\subseteq B$.
Claim: We need to prove $A\cup B\subseteq B$ and $B\subseteq A\cup B$
Let  $x\in A\cup B$. Then  $x\in A$ or $x\in B$. But $A\subseteq B$. Hence $x\in B$. Therefore $A\cup B\subseteq B$.
Now conversely $x\in B$. Then $x\in A\cup B$. Therefore $B\subseteq A\cup B$. Thus $A\cup B=B$.
Consider $A\cup B=B$. 
Claim: $A\subseteq B$
Let $x\in A$. Then  $x\in A\cup B=B$. Therefore $A\subseteq B$.