§ General statement:

Given an nD simplex that has been subdivided, and a function that maps vertices of the subdivision to vertices of the simplex such that the function on the boundary of the simplex maps it to endpoints of the boundary, we will always have a subdivided simplex with all vertices of the original simplex.

§ 1D

• Given a line with endpoints $a, b$ and points in between, we will always have an occurrence of $ab$ on the line.
• Can prove something slightly stronger: there will always be an odd number of $ab$ on the line.

§ 2D

• Given a triangle labelled $abc$ and a subdivision of it, there will be a smaller triangle labelled $abc$.
• Consider all smaller triangles.
• Call a side with $bc$ a door.
• How many doors can a triangle have? It can have 0 doors if it is labelled $aaa$, or $abb$, or some such.
• It can have 1 door if it is:

 a/ \b==c

• It can have two doors if it is:

  c// \b===c

• We can't have three doors. So triangles can have 0, 1, or 2 doors.
• If we find a triangle with one door, we are done, since it will have $abc$.
• Now start from the bottom of the triangle where we have the side $bc$. Here we will find at least one edge $bc$.
• Walk along the triangle, entering any triangle with a door.
• If that's the only door of the triangle, we are done.
• If not, then the triangle has two doors. Exit the current triangle through the other door (the door we did not enter from). This will take us to another triangle.
• See that we cannot terminate the walk by exiting from sides $AB$ or $AC$, for such a side will be of the form $ab$. Then to have a door, we will get a $bc$, so the triangle must be $abc$, ie a triangle we are looking for!
• So if we ever escape the simplex, we must escape from the bottom side $BC$. This removes an even number of $bc$ edges from $BC$. But we know there are an odd number of $bc$ from $BC$, so we must find a triangle $ABC$ eventually.