How to simply prove that a certain graph is a subdivision of $K_{3,3}$

Let $G$ be a graph with six degree $3$ and some degree $2$ vertices and such that every cycle in $G$ contains an even number greater than two of degree $3$ vertices. One must prove that $G$ is a subdivision of $K_{3,3}$.

So, one way of doing this is algorithmically: take a vertex of degree three and follow the three paths pointed by it's three neighbors until you find three new degree $3$ vertices. By the cycle condition you will have that the four degree $3$ vertices you now have found are, in fact, all distinct, and any two of the three new ones can't be connected by a direct path, so all three will be connected by direct paths to the two remaining-to-find degree $3$ vertices (here, a direct path means a path with only degree $2$ internal vertices).

It's easy to see that the graph generated at the end is, in fact, a subdivision of $K_{3,3}$. The problem is that the previous part is far too messy to formalize concisely. I'm having a lot of trouble on writing it. I'm not very advanced in the field, so I can't skip many non-trivial steps that are not known theorems.

I'm looking for a more elegant and direct way to prove this statement. Such a proof can use directly the fact that $K_{3,3}$ is the only graph with six vertices, all with degree $3$, and without odd sized cycles. If you just find a short way to formalize the above demonstration, that would do wonderfully too.

Solution 1:

Repeat the following procedure: for each vertex of degree $2$, delete it and connect its neighbors directly. This "undoes" the subdivision operation.

At the end of this procedure, what we get is potentially a multigraph with $6$ vertices, all of degree $3$. However, the conditions that we have here tell us that it is actually a simple graph:

1. The graph cannot have loops - those are cycles containing only one vertex, which would have been cycles with only one degree-$3$ vertex in the original graph.
2. The graph cannot have multiple edges between the same vertices - these are cycles containing two vertices, which would have been cycles with only two degree-$3$ vertices in the original graph.
3. Also, it is bipartite - any odd cycle would be a cycle with an odd number of degree-$3$ vertices in the original graph.

The only bipartite $3$-regular graph on $6$ vertices is $K_{3,3}$.