## Return to Answer

3 added 16 characters in body

Here is a proof under the assumption that the vertices $V$ are the vertices of a convex polygon. Label these as $p_1, \dots, p_n$ in cyclic order (the subscripts should be read modulo $n$). If all the edges $p_ip_{i+1}$ are red, then we are done. Otherwise, we may assume that $p_1p_2$ is red, and $p_n p_1$ is blue. By induction, we have that $V-p_1$ has a monochromatic spanning red or blue plane tree $T$. In either case, we can extend $T$ to a spanning monochromatic plane tree of $V$.

2 deleted 62 characters in body

Here is a proof under the assumption that $x \notin conv(V-x)$, for all $x \in V$. So, the vertices of $G$ V$are the vertices of a convex polygonin the plane. Label these as$p_1, \dots, p_n$in cyclic order (the subscripts should be read modulo$n$). If all the edges$p_ip_{i+1}$are red, then we are done. Otherwise, we may assume that$p_1p_2$is red, and$p_n p_1$is blue. By induction, we have that$V-p_1$has a monochromatic plane tree$T$. In either case, we can extend$T$to a monochromatic plane tree of$V$. 1 Here is a proof under the assumption that$x \notin conv(V-x)$, for all$x \in V$. So, the vertices of$G$are the vertices of a convex polygon in the plane. Label these as$p_1, \dots, p_n$in cyclic order (the subscripts should be read modulo$n$). If all the edges$p_ip_{i+1}$are red, then we are done. Otherwise, we may assume that$p_1p_2$is red, and$p_n p_1$is blue. By induction, we have that$V-p_1$has a monochromatic plane tree$T$. In either case, we can extend$T$to a monochromatic plane tree of$V\$.