Proof that $A \subseteq B \Leftrightarrow A \cup B = B$

I want to proof that:

$A \subseteq B \Leftrightarrow A \cup B = B$

At the moment I have no idea on how to start. Please give me a hint.

Thx in advance!

A good place to start is always with definitions. Recall the definition of $A\subseteq{B}$:

$$x\in{A}\to{x}\in{B}$$

Now recll the definition of union:

$$A\cup{B}=\{x:x\in{A}\lor{x}\in{B}\}$$

And finally, rewrite the predicate:

$$A\cup{B}=B\leftrightarrow\{x:x\in{A}\lor{x}\in{B}\}=\{x:x\in{B}\}$$

Since you ask about how to start, here is how you can do that:

(1) Assume first that $A\subseteq B$. Then try to prove that $A\cup B = B$. That you can do by first proving that if $x\in A\cup B$ then $x\in B$. And after that, you prove that if $x\in B$ then $x\in A\cup B$ (this is obvious).

(2) Now assume that $A\cup B = B$. Then try to prove that $A\subseteq B$. This you can do by proving that if $x\in A$ then $x\in B$.

First, $A \cup B = B$ means $A \cup B \subseteq B$ and $A \cup B \supseteq B$. However, the latter is trivial, and so we will prove something stronger:

$$A \subseteq B \Longrightarrow A \cup B \subseteq B \tag{1}$$ $$A \subseteq B \Longleftarrow A \cup B \subseteq B \tag{2}$$

where $(1)$ follows from the monotonicity of $\cup$,

\begin{align} A &\subseteq B \\ B &\subseteq B \\ A \cup B &\subseteq B \cup B \end{align}

and $(2)$ is implied by transitivity of $\subseteq$

$$A \subseteq A \cup B \subseteq B.$$

I hope this helps $\ddot\smile$