How can you infer that $A\cap B = \emptyset$? Given:  
$$H:((A \cup B) \to \{ 0,1\} ) \to ((A \to \{ 0,1\} ) \times (B \to \{ 0,1\} ))$$
$$H = \lambda f \in (A \cup B) \to \{ 0,1\} .\left\langle {\lambda a \in A.f(a),\lambda b \in B.f(b)} \right\rangle$$
Based on @Brian M. Scott answer we know that $H$ is injective, but not surjective, and therefore, $H$ isn't invertible. 
How does it tell us that $A\cap B = \emptyset$?

See the comment on the original question :

if $H$ is invertible and the arguments of $H^{-1}$ are $f_A$ and $f_B$, then for $f$ to exists we need $f_A$ and $f_B$ to agree at $A \cap B$, that is $\forall x \in A \cap B$, $f_A(x) = f_B(x)$.

But the definition of Brian Scott couple $<f,g>$ gives you , for $c \in A \cap B$ that :

$f(c) = 1$ and $g(c) = 0$.

