I'm proving that if $A , B\subset S$ then
$A\subset B \Leftrightarrow A\cup B=B$ $A\subset B \Leftrightarrow A\cap B=A$
I go as follows:
Suppose it is true that $A\subset B$. This implies that if $x\in A$ then it is also true that $x \in B$, this is to say
$\tag{1} x\in A\Rightarrow x \in B$
This means that, using the set builder notation
$B=\{ x:x\in B\}=\{x:x\in B \vee x\in A\}=A\cup B$
Now suppose $A\cup B=B$ is true. This means that
$A\cup B=B=\{ x:x\in B\}=\{x:x\in B \vee x\in A\}$
But then this means that if $x\in A$ then it is also true that $x\in B$, which means that $A\subset B$.
(The proof has been now corrected.)