For any two sets how to prove: (1) $A ⊆ B$ (2) $A ∪ B = B$ (3) $A − B = B − A$ I'm trying to prove these statements are equivalent and I'm stuck at proving the second one. I started by proving that  $A \cup B \subseteq B$.
$x \in (A \cup B)  $ then $ x \in B$
$x \in A$ or $x \in B$ then $ x \in B$
$ x \in A $ then $x \in B$ is true from the first assumption
$x \in B$ then $x \in B$ is true because B is a subset of itself
This was in the first direction. ($A \cup B \subseteq B$) But when I try to do it in the other direction I get this.
$B \subseteq A \cup B$
$x \in B$ then $x \in (A \cup B)$
$x \in B$ then $x \in A$ or $x \in B$
I can prove that $x \in B$ then $x \in B$ again but how do I prove $x \in B$ then $x \in A$? ($B \subseteq A$)
 A: Let $A, B$ be any sets.
$(1) \implies (2).\quad$ Suppose that $A \subseteq B.$ Let $x \in A \cup B.$ Then
$$\begin{align}
x \in A \cup B & \implies x \in A \vee x \in B\\
& \implies x \in B \vee x \in B & \text{($A \subseteq B$)}\\
& \implies x \in B.
\end{align}$$
Hence, $A \cup B \subseteq B.$ Now, let $y \in B.$ Then
$$\begin{align}
y \in B & \implies y \in A \vee y \in B\\
& \implies y \in A \cup B.
\end{align}$$
Hence $B \subseteq A \cup B.$ Therefore $A \cup B = B.$
$(2) \implies (3)? \quad$ No. Let $A = \{1\}$ and $B = \{1,2\}.$ Note that $A \cup B = \{1,2\} = B$ But $A - B = \emptyset \neq \{2\} = B -A.$
$(3) \implies (1). \quad$ Suppose that $A - B = B - A.$ Then $A = B.$ Therefore $A \subseteq B.$
$(2) \implies (1). \quad $ Suppose that $A \cup B = B.$ Let $x \in A.$ Then
$$\begin{align}
x \in A & \implies x \in A \vee x \in B\\
& \implies x \in A \cup B\\
& \implies x \in B & \text{($A \cup B = B$)}.
\end{align}$$
Therefore $A \subseteq B.$
$(3) \implies (2). \quad$ Suppose that $A \cup B = B.$ Hence $A = B.$ Then
$$A \cup B = B \cup B = B.$$
$(1) \implies (3)? \quad$ No. Use the same example used in proving that $(2) \implies (3)$ is false.
So the implications $(1) \implies (2), (3) \implies (1), (2) \implies (1)$ and $(3) \implies (2)$ are true, while $(2) \implies (3)$ and $(1) \implies (3)$ are false.
More, you can see that $(1) \iff (2)$ so $(1)$ and $(2)$ are the only equivalent statements. $\quad \square$
