The question/task is to prove/disprove the conjecture below.
Let $G$ be a maximal planar graph with a 4-coloring $f$. Let $(a,b,c,d)$ be a cycle in $G$. Let $S$ be the collection of all $a,c$-paths in $G$ and all $b,d$-paths in $G$.
Conjecture: At least two members of $S$ are bicolored.
(i.e., there exist distinct paths $Q_1,Q_2\in S$ and colors $i,j,k,l\in\{1,2,3,4\}$ such that $f(u)\in\{i,j\}$ for every vertex $u$ on $Q_1$ and $f(v)\in\{k,l\}$ for every vertex $v$ on $Q_2$).
Definitions:-
$G$ is a maximal planar graph if it can be drawn on a plane such that no edges cross and boundary of every face is a triangle. A 4-colouring $f$ of $G$ is a function $f:V(G)\to\{1,2,3,4\}$ such that $f$ map endpoints of each edge to different 'colors' (i.e. $f(u)\neq f(v)$ for every edge $uv$ of $G$).
Notes:-
It is easy to see that the conjecture is true if the cycle $(a,b,c.d)$
is tricolored (or bicolored). The following is the crux of the conjecture.
Let $G$ be a planar graph with a 4-coloring $f$. Let $(a,b,c,b)$ be a cycle in $G$ such that boundary of each face inside the cycle is a triangle. Suppose that the cycle $(a,b,c,d)$ receives all four colors.
Conjecture 2: Then, there is a bicolored $a,c$-path or a bicolored $b,d$-path inside the cycle.
If Conjecture 2 is true, then the main conjecture above is true (apply Conjecture 2 to outside region of the cycle $(a,b,c,d)$ ).