Understanding Rosenthal 3.1.5 (i) 
If $X = 1_A$ is the indicator of some event $A \in \mathcal{F}$, then $X$ is a random variable.

The proof for this states $X^{-1}(B)$ (for $B$ being in all Borel sets $\mathcal{B}$) must be either $A, A^c, \emptyset, \Omega$.  But I do not think this is so clear, so proof:
Let $X = 1_{A}$ for $A \in \mathcal{F}$; that is for $A \in \mathcal{F}$, and $\mathcal{F}$ is a sigma algebra
$\forall \omega \in \Omega: X(\omega) = 1_{A} (\omega) =
\begin{cases}
1 & if \omega \in A \\
0 & if \omega \notin A \\
\end{cases}
$
Then the pre-image of $X$ is:
$\forall B \in \mathcal{B}: X^{-1}(B) = 
\begin{cases}
A & if \; 1 \in B \\
A^c & if \; 0 \in B \\
\emptyset & if \; \{ 0, 1 \} \not\in B
\end{cases}
$
But there is no $\Omega$ here.  I have seen this proof which evaluates at $x \in \mathbb{R}$, but here we are evaluating on the Borel sets which may be intervals.  So the pre-image of the interval $(2,3)$ will be empty, but will $X^{-1} (1/2,3/2) = A$ or can only $X^{-1} (1) = A$ ?

There is a problem with your calculation of the preimage. In fact
$$X^{-1}(B) = 
\begin{cases}
A & if \; B \cap \{0,1\} = \{1\} \\
A^c & if \; B \cap \{0,1\} = \{0\} \\
\emptyset & if \; B \cap \{ 0, 1 \} = \emptyset \\
\Omega & if \; B \cap \{0,1\} = \{0,1\}
\end{cases}$$
