Note that if the numbers assigned to all vertices are equal, then Rudi can choose any diagonals.
Hence we assume there is at least one $0$-vertex and at least one $1$-vertex. (We shall call the vertices "0-vertex" or "1-vertex", according to the number assigned to it.)
By the strong induction on the number $n$ of vertices we will prove that Rudi can choose diagonals in such a way that the number written in each triangle is either $1$ or $2$.
If $n = 3$, the statement is obviously true.
The case $n = 4$ is also easy. If there is a diagonal connecting a $0$-vertex with a $1$-vertex, then Rudi draws that diagonal. Otherwise he draws any diagonal and obtains equal numbers ($1$ or $2$) in both triangles.
Assume that the statement is true for all polygons with less than $n$ vertices ($n \ge 5$), such that there are both a $0$-vertex and a $1$-vertex.
In a polygon with $n$ vertices, we can choose two adjacent vertices $A$ and $B$ so that $A$ is a $0$-vertex and $B$ a $1$-vertex. Let $P$ be a vertex which is not adjacent to $A$ or $B$ (such vertex exists since $n \ge 5$). If $P$ is a $0$-vertex, Rudi connects it to $B$, otherwise to $A$. In this way Rudi has divided the polygon into two polygons with smaller number of vertices and each of them has both a $0$-vertex and a $1$-vertex. By the inductive hypothesis each of the two smaller polygons can be divided into triangles such that the numbers $1$ and $2$ are written in all of them, so the same holds for the polygon with $n$ vertices. This finishes the inductive step.