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 $A\cap B = \emptyset$?

Assume for a contradiction that $k \in A \cap B$ and that $H$ is surjective. Then $H(K) = \langle \{k\}_A,\emptyset_B\rangle$ for some $K : A \cup B \rightarrow 2.$ Fix any such $K$.
Case 1. Assume $k \in K$. Then $k \in [H(K)]_0$ and $k \in [H(K)]_1$, thus $H(K) \neq \langle \{k\}_A,\emptyset_B\rangle,$ a contradiction.
Case 2. Assume $k \notin K$. Then $k \notin [H(K)]_0$ and $k \notin [H(K)]_1$, thus $H(K) \neq \langle \{k\}_A,\emptyset_B\rangle,$ a contradiction.
In conclusion: if $H$ is surjective, then $A \cap B$ must be empty.
