# $2-$cell embedding of a graph in a surface

I want to prove that every graph $$G$$ embeddable in the projective plane has a vertex of degree $$\leq 5$$. If I suppose the graph has a $$2-$$cell embedding (an embedding in a way every face is homeomorphic to a disk) we can extend it to a triangulation and in a triangulation we can easily prove using Euler's formula that there are exactly $$3n-3$$ edges, from which we get that $$\delta(G)\leq 2\frac{3n-3}{n}<6$$ so we get the result.

In the general case, when the embedding is arbitrary, I wanted to approach it proving that every graph which can be embedded in the projective plane, can be seen as a subgraph of a graph with a $$2-$$cell embedding. I don't know if this is true, but at the least, if the graph is planar, we can add edges to make it into a maximal planar graph, so that the outer face (in a plane embedding) is a triangle, and drawing it inside the cycle $$4564$$ in the following $$2-$$cell embedding of $$K_{6}$$ (edges of the same color are identified) would make it a subgraph of a $$2-$$cell embeddable graph.

I think the same approach would work to prove that planar graphs can be drawn as subgraphs of $$2-$$cell embeddable graphs in any compact and closed surface.

But what happens if the graph is not planar? Is there a better approach for this?