Coupling of r.v. I am trying to answer this question.

If $X$ and $Y$ are random variables on $(\Omega, \mathcal{B})$, show 
  \begin{align*}
\sup_{A \in \mathcal{B}} |P[X \in A] -P[Y \in A]| \le P[X \neq Y].
\end{align*}

Things I tried.
\begin{align*}
|P[X \in A] -P[Y \in A]|&=P \{[X \in A] \Delta [Y \in A] \}\\
&=P\{ \omega: \left(X^{-1}(A)\backslash Y^{-1}(A)\right) \cup \left(Y^{-1}(A) \backslash X^{-1}(A) \right) \}\\
&=P\{ \omega: X^{-1}(A)\backslash Y^{-1}(A) \} + P \{Y^{-1}(A) \backslash X^{-1}(A) \}
\end{align*}
On the other hand 
\begin{align*}
 P[X \neq Y]&=P \{\omega: X^{-1}(\mathcal{B}(\mathbb{R})) \Delta Y^{-1}(\mathcal{B}(\mathbb{R})) \}\\
&=P \{\omega: \left( X^{-1}(\mathcal{B}(\mathbb{R})) \backslash Y^{-1}(\mathcal{B}(\mathbb{R}))\right) \cup \left( Y^{-1}(\mathcal{B}(\mathbb{R})) \backslash X^{-1}(\mathcal{B}(\mathbb{R}))\right)   \}\\
&=P \left\{\omega: \left( X^{-1}(\mathcal{B}(\mathbb{R})) \backslash Y^{-1}(\mathcal{B}(\mathbb{R}))\right) \right\} + P \left\{ \left( Y^{-1}(\mathcal{B}(\mathbb{R})) \backslash X^{-1}(\mathcal{B}(\mathbb{R}))\right)   \right\}
\end{align*}
Assuming all I did is correct. Next, I think, I must use definition of $\sup$ to show the rest. I feel like I am almost there. Can any one help? Thanks. 

In general if $P$ is a probability and $R,S$ are measurable sets:
$\left|P\left(R\right)-P\left(S\right)\right|=\left|P\left(R\cap S^{c}\right)-P\left(R^{c}\cap S\right)\right|\leq P\left(R\cap S^{c}\right)+P\left(R^{c}\cap S\right)=P\left(R\triangle S\right)$
Applying that here gives for every $A\in\mathcal B$:
$\left|P\left\{ X\in A\right\} -P\left\{ Y\in A\right\} \right|\leq P\left(\left\{ X\in A\right\} \triangle\left\{ Y\in A\right\} \right)\leq P\left\{ X\neq Y\right\} $
The second inequality as a consequence of $\left\{ X\in A\right\} \triangle\left\{ Y\in A\right\} \subseteq\left\{ X\neq Y\right\} $
