Prove that $\bigcup \{A,B \} = A \cup B$. 
Prove that $\;\bigcup\big\{A,B\big\} = A \cup B$.
I am trying to work on this problem here. But I just do not know where to start.
So I know that the union of set {A,B} is {A,B}. And the union of Set A and B we also get {A,B} for any common elements. How would I show this proof wise
 A: Let $x \in \bigcup \{ A, B \}$ be arbitrary. This means there exists $C \in \{ A, B \}$ such that $x \in C$. We can simply enumerate to find that either $x \in A$ or $x \in B$, and this implies $x \in A \cup B$. So $\bigcup \{ A, B \} \subseteq A \cup B$.
Let $x \in A \cup B$ be arbitrary. This means either $x \in A$ or $x \in B$. If $x \in A$, then because $A \in \{ A, B \}$, it must be that $x \in \bigcup \{ A, B \}$. If $x \in B$, then because $B \in \{ A, B \}$, still $x \in \bigcup \{ A, B \}$. So $A \cup B \subseteq \bigcup \{ A, B \}$.
Therefore, $\bigcup \{ A, B \} = A \cup B$.
A: 
So I know that the union of set {A,B} is {A,B}.

It seems like you are not clear on what $\bigcup \mathcal S$ means when $\mathcal S$ is a set of subsets of $X$.  The definition is:
$$
\bigcup \mathcal S=\{x\in X\mid \exists S\in\mathcal S \text{ such that } x\in S\}
$$
So while $\bigcup\{\{A,B\}\}=\{A,B\}$, it is not true that "$\bigcup\{A,B\}=\{A,B\}$."
I recall the helpful thing someone said to me the first time I was learning it: they said "it's like the union symbol breaks down the walls between sets in $\mathcal S$."
So, for example, $\bigcup \{\{a,b\},\{c,d\}\}$ dissolves the inner braces and you just get $\{a,b,c,d\}$.
As far as I know, $A\cup B$ is just a convenient shorthand for $\bigcup \{A,B\}$ in order to think of it as a binary operator.  I'm not aware of any definition that defines the two differently from each other.
