The best you can do is 9. Let's number corners as 1-4 on top, 5-8 on bottom (
i is connected to
i+4). You can do:
(1-2), (2-3), (3-4), (4-1), (1-5), (5-6), (6-7), (7-8), (8-5)
which has 9 edges.
Why is this the best?
As the cube is a graph with 12 edges and 8 vertices and each vertex has a degree of 3. Effectively this means that passing through a vertex "removes" 2 of that degree. The other 1 must be an end of a sequence of vertices and edges. Having 8 "ends" means that you must have 4 sequences of vertices and edges.
As each sequence has at least one edge, by minimizing three of those sequences to a length of one edge, we get that the best we can do is 9 for the fourth sequence.
For a more theoretical approach to that problem have a look at this.