Prove that there is an even number of faces whose vertices get all three labels in a graph that is triangular Consider a planar drawing of a graph all whose faces, including the
outer one, are triangular (i.e., have 3 vertices). To each vertex, we arbitrarily assign one of the labels 1,2,3. Prove that there is an even number of faces whose vertices get all three labels.
I'm trying to prove the above and I got very close with my intuition but I'm not yet there. This is what I did:

I assigned odd degree with having all the three vertices. The red one is the only one that has all of them and the outer one i guess. We know that it can't only be the red one because by handshake lemma we know that the number of vertices with an odd degree is even but is this a good proof? And can I assign stuff like I did?
 A: You are on the right track.
First, in your example the outer face is not triangular. 
Second, you cannot just assign odd degrees, what you should do, is to define a graph in such a way that the right vertices will have odd degrees.
So let's define a (multi)graph $G'$ that will have faces of $G$ as vertices, and there will be an edge between two vertices $f_1$ and $f_2$ of $G'$ (i.e., faces of $G$) when:


*

*Faces $f_1$ and $f_2$ share an edge in $G$.

*Labels on the endpoints of the shared edge in $G$ are different.


Technically we should make $G'$ a multigraph, because two faces can share more than one edge. With such a definition, indeed the vertices of $G'$ (which are faces of $G$) have an odd degree if and only if all three labels different (degree zero if there is only one unique labels and degree two if there are two unique labels).
Finally, if you are uncomfortable using faces of $G$ as vertices of $G'$, you can map them to numbers, that is, if $G$ has $F \in \mathbb{N}$ faces, then we will number them arbitrarily as $f_1, f_2, \ldots, f_{F}$ and then construct graph $G'$ on the set $\{1, 2, \ldots, F\}$. In this way there is a one-to-one correspondence between the vertices of $G'$ and faces of $G$ and so the whole reasoning works just as well. See also the definition of a dual graph (e.g., $G'$ is a subgraph of the dual of $G$).
I hope this helps $\ddot\smile$
