Ant cannot travel to opposite side of cube in even amount of steps Say an ant starts at vertex 1 of a cube and wants to travel to vertex 8. The ant cannot do this in 4 steps, nor any even amount of steps in general. How can I express this? It seems that after every first move of one, the number is guaranteed to be odd because there will be an even amount of moves made after the additional first move.
The cube graph is an example of a bipartite graph, where the vertices can be  divided into two group such that the edges always connect a vertex from one group to a vertex in the other group, never to another vertex in the same group. 

The group can be seen in your diagram by moving vertex $1$ in amongst the other odd-numbered vertices and vertex  $8$ likewise into the even-numbered group. Then you will always traverse an odd number of edges to reach a vertex in the other group from the starting vertex, plus all cycles are even length, etc.
As you can see you have bipartite graph here and one of the things of bipartite graph is that for any u,v vertices all the paths from u to v has to odd or even (can't be 1 path odd while there is other even) so if we have bipartite graph and we show 1 odd path then all the paths are odd.