Use Direct Proof to Prove: If $A\cup B = A$, then $B\subseteq A$ I started with:

Assume $A\cup B = A$, then $x\in A \cup B$. 
Without loss of generality, let $x \in A$.

However, at this point I am not sure what to do. 
Note: In the textbook, they use proof by contrapositive, but I want a direct proof.
 A: Along the lines of @T.Bongers's comment:

Let $x \in B$. 
Then $x \in A\cup B$ (by definition of union).
By hypothesis $A \cup B=A$.
Therefore $x \in A$.
Since $x$ was arbitrary, we have shown that for any $x \in B$, $x$
  is also an element of $A$.
It follows that $B \subseteq A$ by the definition of subset.

A: $\{ x: x \in A \ \text{or}  \ x \in B\} = \{ x: x \in A \}$, so $x  \in A \cup B \Longrightarrow x \in A$. 
Thus, $\{x \in A \Longrightarrow x \in A\} \ \land \ \{x \in B \Longrightarrow x \in A\} $ implies $B \subseteq A$
A: $A\cup B=A$ iff there does not exist an element of $B$ not in $A$ iff $B$ is a subset of $A$. 
A: If $A\cup B=A$ , then every element of the union of $A$ and $B$ is an element of $A$.   Now every element of $B$ is something because of reasons, so therefore the assumption infers that every element of $B$ would be an element of $A$.  $$A\cup B=A \implies B\subseteq A$$
$\Box$

More directly: Because of some statement by definition, then if $A\cup B=A$ is assumed we may infer $B\subseteq A$.
$\Box$
