Take any triangulation of a square (= a partition into triangles such that any two triangles are either disjoint or intersect in a common face). Suppose that there are no triangulation vertices on the boundary of the square, except the four corners.
Color each vertex either red or blue, with the following constraints:
- Two opposite corners are red and two opposite corners are blue;
- In every triangle, at least one corner is red and one corner is blue.
Here is an example:
In all examples I checked, there is either a red path between the two red corners, or a blue path between the two blue corners. In the example above there is a blue path $D_1-D_4-D_6-D_2$; if we change e.g. $D_4$ to red, then there is a red path $C_1-C_3-C_4-D_4-C_8-C_2$.
Does there always exist either a red path between the red corners or a blue path between the blue corners?
NOTE: without the condition that there are no triangulation vertices on the boundary, the claim is false: consider the inner square with corners $C_7,D_7,C_8,D_8$.
Note: The assumption that every triangle has a red and blue vertex is superfluous. As long as one pair of opposite external vertices is red, and the other is blue, then one of the pairs will be connected by a monochrome path.
Your problem is a generalization of proving Hex always has a winner, so I adapted a well-known proof of the Hex problem to your problem.
Imagine that every edge with a red and blue endpoint is a door, and every edge with same-colored endpoints is a wall. Every triangular cell is a room with exactly two doors. (Without your second bullet assumption, you could still say every room has either $0$ or $2$ doors.)
Start by entering the door between $C_1$ and $D_1$. Every time you enter a triangle via a door, exit via the other door. This process can only stop when you exit via one of the other edges of the square.
If you exit via the $(C_2,D_1)$ edge, then the union of the set of red vertices on the rooms you visited will be a connected set containing $C_1$ and $C_2$, so a red path exists.
If you exit via the $(C_1,D_2)$ edge, the same logic implies a blue path exists.
It is impossible to exit via the opposite $(C_2,D_2)$ edge. This would imply that both a red and blue path existed, but two such paths would have to cross (by the Jordan curve theorem, say).
To be a bit more precise, define a sequence of red vertices $r_1,r_2,r_3,\dots$, and a sequence of blue vertices $b_1,b_2,b_3,\dots,$ as follows. Start with $r_1=C_1$, $b_1 = D_1$. Suppose that $r_n$ and $b_n$ have been determined. If $n=1$, there is a unique triangle $T$ with $r_1$ and $b_1$ as vertices. If $n\ge 2$, there are two such triangles, and we define $T$ to be the one which was not considered in the previous step. Let $v$ be the third vertex of $T$. If $v$ is red, then we define $(r_{n+1},b_{n+1})=(v,b_n)$. If $v$ is blue, we define $(r_{n+1},b_{n+1})=(r_n,v)$.
The process ends once $(r_n,b_n)$ is one of the other external edges, in which case the only triangle with $(r_n,b_n)$ as an edge would be the triangle considered in the previous step. If the edge you exit by contains $C_2$, then $(r_1,r_2,\dots,r_n)$ is a walk from $C_1$ to $C_2$. If the edge you exit by contains $D_2$, then $(b_1,b_2,\dots,b_n)$ is a walk from $D_1$ to $D_2$.
This generalizes to any number of dimensions. Instead of a square, consider a triangulation of the $n$-orthoplex. This has $2n$ vertices of the form $\{\pm e_1,\dots,\pm e_n\}$, where $e_i$ is the $i^\text{th}$ standard basis vector. As before, we color the vertices of this triangulation in $n$ colors, such that both $e_i$ and $-e_i$ get the $i^\text{th}$ color. You can conclude that one of the opposite pairs, $\{\pm e_i\}$, will be connected by a path whose vertices are all colored $i$. The proof is analogous; imagine that each $(n-1)$-dimensional face is a door provided that its $n$ vertices span all $n$ colors. Enter the door spanned by $\{e_1,\dots,e_n\}$, and whenever you enter a simplex, leave by the unique other door. When you exit the orthoplex, you will have found your path.