Prove a statement with elements for Set Theory I am stuck on this proofing question and I would like some clarification.
Q: $A\subseteq B \iff A\cap B^{\prime} = \emptyset$
I already proved that LHS goes to RHS, but I am confused for the other way around because the textbook answer key gives a weird answer.
It says that for $A\cap B^{\prime}=\emptyset$, let $x$ be an element of $A$. If x isn't an element of $B$ then, then $x$ is an element of $B^{\prime}$, therefore $x$ is an element of $A\cap B = \emptyset$. Hence $x$ is an element of $B$ and $A$ is a subset of $B$.
I am mainly confused about how they say $x$ isn't an element of $B$ and then all of a sudden say $x$ is an element of $B$...? How could it be both an element and not an element of $B$??

The point is that there's a contradiction here; this is how a standard proof by contradiction goes: Start by assuming something (that is hopefully false), and use it to get something you know is false. Then the original statement must be false too.
So to clarify the proof, I'll expand it a bit:

Suppose, intending a contradiction, that $x \in A$ but $x \notin B$. However, this implies that $x \in B^c$, so that $$x \in A \cap B^c = \emptyset$$
  This is the desired contradiction, so $x$ couldn't have existed to start with. Thus if $x \in A$, we have $x \in B$ too, so that $A \subseteq B$.

