Sperners Lemma

There are basically two lemmas in Topology which allow us to prove Sperner's Lemma.
Lemma:1 If a close interval [0,1] has finite number of points that divide it into smaller intervals by points either 0 and 1. Then there are odd number of intervals whose end points are different.
Proof: Think of the left most 1 out of all the one. This will form the very first interval of the type (0,1). The next interval if is present will be of type (1,0) and this forces the presence of another interval of type (0,1). So we see that the intervals of type (0,1) occurs at odd positions and since the last interval has to be of type (0,1). It means we will have odd number of intervals.

Lemma:2 Suppose any room of a house has 0,1 and 2 doors. Then the number of dead ends and the outer door have the same parity.
Proof: First thing to notice is that there is a unique path from any given door to another door. Second, there are only 3 such path exist.
1. Outer door to Outer door (let there be m such paths. For each path we have two doors, therefore we have 2m such outer doors).
2. Outer door to Dead end door (let there be n such paths. For each path we have one outer and one dead end door, therefore we have n outer and n dead end doors)
3. Dead end door to Another Dead end door. (let there be p such paths. For each path we have 2 dead end doors. Therefore we have 2p dead end doorss)
Therefore
Number of outer doors = 2m+n
Number of dead end doorss = 2p+n
Thus both have same parity.

Sperners Lemma: It gaurantees the existence of a dead end triangle in any triangulation of a triangle obtained by the following method. The vertices are labelled 1,2 and 3. Then the side containing (1,2) vertex has only 1 and 2 intermediate point and so on. The vertices of triangle inside could be labeled by any of these three numbers.
Proof: Suppose (1,2) and (2,1) are the doors. Then by lemma 2 we have only doors on one side of the triangle and there will be odd number of these. So if we enter through any of the doors either we exit from another door then pairs are taken. So we will always be left by door which is not matched and that mean it will end up going into a triangle with all three different vertices.
Remark: There are only 2 types of triangle which allow a path (1,1,2), (2,2,1) and (1,2,3). All other triangles are inaccessible.

Following two videos explain sperners lemma and using fair division.

This gives a little more explanation to the above. Both complement each other.