I posted this question previously, but now I think I have part of the solution and just need a bit more guidance to finish the proof:
"For any $n \in \mathbb{N}$, let {0, 1, 2}$^n$ = { $(a_1, a_2, ..., a_n)$ | for all $i \in \mathbb{N}_n, a_i \in$ {0, 1, 2} }. Construct a bijection from {0, 1, 2}$^n$ to {$(A,B) | A,B \subset \mathbb{N}_n$ and $A,B$ are disjoint}."
Firstly, I defined the characteristic function $$ \chi(S)= \begin{cases} 0&\text{if}\, S \in A\\ 1&\text{if}\, S \in B\\ 2&\text{if}\, S \in (\mathbb{N}_n-A-B)\\ \end{cases} $$ to map from $\mathbb{N}_n$ to {0,1,2}, where $S$ is a subset of $\mathbb{N}_n$. That is, any particular characteristic function will determine a specific element $(A, B)$ in the set {$(A, B)$}. Then the set $Fun(\mathbb{N}_n$, {0, 1, 2}$)$ contains all characteristic functions, determining the entire set {$(A, B)$}.
At this point, I don't know how to proceed to actually create a bijection from {0, 1, 2}$^n$ to {$(A, B)$}.
The set of pairs of disjoint subsets of $\Bbb N_n$, I will denote $\mathcal{P}$, say. Your function $\chi$ is the right idea: it sort of is the bijection, but formally we define a $\chi: \mathcal{P} \to \{0,1,2\}^{\mathbb{N}_n}$ by
$$\forall (A,B) \in \mathcal{P}: \forall x \in \Bbb N_n: \chi((A,B))(x) = \begin{cases} 0 & x \in A\\1 & x \in B\\ 2 & x \notin (A \cup B)\end{cases}$$
This makes $\chi((A,B))$ a well-defined member of $\{0,1,2\}^{\Bbb N_n}$ for $(A,B) \in \mathcal{P}$: $A$ and $B$ are disjoint so any $x$ is in exactly one of the cases in the definition of $\chi((A,B))$, and its values are only in $\{0,1,2\}$ on inspection.
Now we can define an inverse function $\Psi: \{0,1,2\}^{\Bbb N_n} \to \mathcal{P}$ such that $$\forall (A,B) \in \mathcal{P}: \Psi(\chi((A,B))) = (A,B)\tag{1}$$ and $$\forall f \in \{0,1,2\}^{\Bbb N_n}: \chi(\Psi(f))=f\tag{2}$$
$$\forall f \in \{0,1,2\}^{\Bbb N_n}: \Psi(f)=\left(f^{-1}[\{0\}], f^{-1}[\{1\}]\right) \in \mathcal{P}$$
where $f^{-1}[\{i\}]=\{x \in \Bbb N_n: f(x)=i\}$ as usual, for $i=0,1$. Disjointness is clear as the value of an $x$ is not both $0$ and $1$..
And $(1)$ and $(2)$ is just definition tracing. If you start out with some pair, and define $\chi$ as described, $A$ is exactly the set on which that function is $0$ etc. So $\Psi$ applied to $\chi((A,B))$ gives back the pair $(A,B)$ again.
If you're more familiar with it, you can also give proofs of $\chi$ being injective and surjective. But the same ideas then reoccur.