Q:

Set theory exercise involving union

$A\subseteq B$ iff $A\cup B = B$
Attempted proof - Suppose $A\subseteq B$ and $x\in A\cup B$ then $x\in A$ or $x\in B$ or both. If $x\in A$ then by hypothesis $x\in B$ as well since $A\subseteq B$, same applies for if $x\in B$. Then this implies that $A\cup B\subseteq B$. 
I have no idea where to go from here. Any suggestions is greatly appreciated.

A:

So you have proved that "$A \subseteq B$ and $x \in A \cup B$ implies $x \in B$". You still need to prove that "$A \subseteq B$ and $x \in B$ implies $x \in A \cup B$" (which will complete the first direction of the equivalence), and then "$A \cup B = B$ implies $A \subseteq B$" (which is the other direction).
To reiterate your first direction:

If $A \subseteq B$, we show that $A \cup B = B$. It is enough in fact to show that $A \cup B \subseteq B$, because $B \subseteq A \cup B$ is always true.
So suppose $x \in A \cup B$; we need to show that $x \in B$. Since $x \in A \cup B$, we have either $x \in A$ or $x \in B$.

If $x \in B$, we are immediately done.
If $x \in A$, then by the hypothesis that $A \subseteq B$, we have $x \in B$, so we are done again.

By this point we have proved that $A \subseteq B$ implies $A \cup B = B$.
For the converse, suppose $A \not \subseteq B$.
Then there is $x \in A$ which is not in $B$.
But then $x \in A \cup B$, and $x \not \in B$, so it is not the case that $A \cup B = B$: we have exhibited something which is in one but not the other.

