To begin, notice that the diagonals of $\Omega$ divide the polygon into $C_n^4 + C_{n-1}^2$ small regions. Every quadrilateral $P_iP_jP_kP_l$ ($1 \le i < j < k < l \le n$) has a 1-1 correspondence with the intersection of the diagonals $P_iP_k$ and $P_jP_l$, and additionally, every intersection is adjacent to four small regions. Therefore, it is enough to assign every intersection to one of the four adjacent small regions, such that the $C_n^4$ assigned small regions are all distinct.

Place $\Omega$ in the Cartesian coordinate plane such that no diagonal is parallel to the $x$ axis. If two diagonals $P_iP_k, P_jP_l$ meet at $Q$, then we assign $Q$ to the unique small region incident to $Q$ and whose interior points all have $y$ coordinates larger than that of $Q$ (intuitively, above $Q$). Obviously, different intersections correspond to different small regions, as for each small region $R$, there is a unique vertex with the minimum $y$ coordinate, and this vertex is assigned to $R$. This completes the proof. $\square$

As in the first proof, we are required to assign each intersection (of two diagonals) to an adjacent small region, such that the assigned small regions are all distinct. In the plane of $\Omega$, pick any point $Q$ not concurrent with any diagonal. Suppose that two diagonals $\ell_1$ and $\ell_2$ intersect at $R$. Then $Q$ lies inside one of the four zones of the plane divided by $\ell_1$ and $\ell_2$: in the anticlockwise direction, let $T$ be the next zone, and assign $R$ to the unique small region adjacent to $R$ and contained in $T$. It remains to prove that each small region is assigned at most one intersection.

Fig. 1.1

For a small region illustrated in Fig. 1.1, denote its vertices anticlockwise as $R_1, R_2, \dots, R_m(R_0 = R_m)$. The rays $R_1R_2, R_2R_3, \dots, R_{m-1}R_m, R_mR_1$ divide the plane excluding the small region into $A_1, A_2, \dots, A_m$, where $A_i$ is the zone whose boundary consists of rays $R_iR_{i+1}$ and $R_{i-1}R_i$. According to the correspondence between small regions and intersections, this small region is assigned $R_i$ if and only if $Q$ lies inside zone $A_i$, which could happen for at most one subscript $i$. Since this is true for any small region, we infer that each small region is assigned at most one intersection, and the proof is completed. $\square$

Let all the diagonals of $\Omega$ be $\ell_1, \dots, \ell_m$. We put these $m$ diagonals in $\Omega$ in the order of the subscripts and define $S_k$ as the set of intersections created by adding the diagonal $\ell_k$ ($S_k$ could be empty), $T_k$ as the set of all small regions divided by $I_1, \dots, I_k$. We use induction to prove, for each $k$, there exists a one-to-one mapping
$$f_k : S_1 \cup \dots \cup S_k \to T_k,$$
such that every $Q \in S_1 \cup \dots \cup S_k$ is a vertex of the polygon $f_k(Q)$. After that, a discussion as in the beginning of the first proof will lead to the conclusion.

For $k=1$, the assertion is obvious. Assume for $k-1$, the one-to-one mapping $f_{k-1}$ satisfies the condition. Now we add the diagonal $I_k$ to $\Omega$.

If $S_k$ is empty, the assertion is clearly true for $k$. Assume $S_k$ is nonempty. Then the intersections in $S_k$ all lie on $\ell_k$, say they are $Q_1, \dots, Q_r$, and they traverse the small regions $A_1, \dots, A_{r+1}$, respectively, dividing each $A_i$ into two new small regions $B_i$ and $C_i$.

Notice that $f_{k-1}^{-1}(A_i)$ can be either empty or determined, say a vertex $R_i$ of $B_i$ or $C_i$. Now we define $f_k$ as follows: for each $i \in \{1, \dots, r\}$,

(1) If $f_{k-1}^{-1}(A_i)$ is empty, then let $f(Q_i) = B_i$;

(2) If $R_i$ is a vertex of $B_i$, then let $f_k(R_i) = B_i$, $f_k(Q_i) = C_i$;

(3) If $R_i$ is a vertex of $C_i$, then let $f_k(R_i) = C_i$, $f_k(Q_i) = B_i$.

For all other intersections $Q$, let $f_k(Q) = f_{k-1}(Q)$. It is straightforward to check that $f_k$ has the desired properties. Particularly, when $k=m$, the mapping $f=f_m$ gives a one-to-one correspondence between all interior intersections and small regions of $\Omega$, satisfying that each intersection $Q$ is a vertex of the small region $f(Q)$. $\square$