Could someone explain this proof to me? Probability proof 
If event $A$ and $B$ are events such that $P(A)$ and $P(B)$ are either $0$ or $1$ and $A$ is subset of $B$, then $A$ and $B$ are dependent events.
Proof: Since $A\subset B$, we have $A\cap B=A$ and so $P(A\cap B)=P(A)$.
$\therefore$ $P(A\cap B)-P(A)P(B)=P(A)-P(A)P(B)=P(A)[1-P(B)]$
Since $P(A)>0$ and $P(B)<1$,

(original image)
Where did that last step come from, involving $P(A \cap B) - P(A)P(B)$? How did the solution come up with that?
 A: The result has clearly been misstated or miscopied. The proof is correct for the following result:

If event $A$ and $B$ are events such that $P(A)$ and $P(B)$ are neither $0$ nor $1$, and $A$ is subset of $B$, then $A$ and $B$ are dependent events.

The reason for looking at $P(A\cap B)-P(A)P(B)$ is that by definition, $A$ and $B$ are independent if and only if $P(A\cap B)=P(A)P(B)$, i.e., if and only if $P(A\cap B)-P(A)P(B)=0$. But the hypothesis that $A\subseteq B$ implies that $P(A)=P(A\cap B)$, so $$P(A\cap B)-P(A)P(B)=P(A)-P(A)P(B)=P(A)\big(1-P(B)\big)\;,$$
which is $0$ if and only if either $P(A)=0$ or $1-P(B)=0$, i.e., if and only if either $P(A)=0$ or $P(B)=1$. These possibilities are ruled out by the corrected version of the hypothesis, so it must be the case that $P(A\cap B)-P(A)P(B)\ne 0$, $P(A\cap B)\ne P(A)P(B)$, and hence by definition $A$ and $B$ are not independent (which of course means that they are dependent).
The answer by copper.hat shows why the stated version is wrong.
A: There are only three possibilities to consider:
$$\begin{matrix}
PA & PB & P (A \cap B) & PA.PB \\
\hline
0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
1 & 1 & 1 & 1
\end{matrix}$$
In all cases $PA.PB = P(A \cap B)$, so the events are independent.
A: Seems like there's something wrong with the initial statement to me.
If $ A = \emptyset \subset B $ then $P(A \cap B)=P(\emptyset \cap B)=P(\emptyset)=0=P(\emptyset)P(B)=P(A)P(B)$.
