Are there graphs embeddable on $S^2$ but not on $\mathbb{R}^2$?

Question

The answer is No for finite graphs, of course, via the stereographic projection. But we need a free point on $S^2$ for that to work. Let $\mathfrak{c}$ denote the cardinality of the continuum.

I can embed the edgeless graph $\overline{K_\mathfrak{c}}$ on $S^2$ by covering each point of it, making a stereographic projection impossible, so there exists embeddings that cannot be transferred to $\mathbb{R}^2$ via the stereographic projection. (Let's point out here that I'm speaking about combinatorical immersions, not topological ones, see here, here and end of post, the example should have made that clear.) But on the other hand I could also embed $\overline{K_\mathfrak{c}}$ on any smaller closed (or open) neighborhood of $S^2$, and I can project.

Another example could be the biparite graph $K_{2,\mathfrak{c}}$, with the two vertices at the poles and the others at the equator, all connected with geodesic lines, covering the whole sphere. But then again I could also just use half the equator and get a free half sphere for the projection.

Another idea would be the dipole graph $D_\mathfrak{c}$ (multigraphs are allowed) covering the sphere, but that can be solved just like the other two examples.

So, my initial guess was that I can't get the graph enough connected to make another easier embedding impossible while keeping it embeddable to the sphere. But I don't know how to show that.

Then I stumbled upon John Sample's answer here, but I doubt both methods provided (blow up a vertex and blowing up an edge (in the comments)), due to a lack of understanding.

• Blowing up a vertex seems to assume that the disk will have "free points" not connected to an edge, but I don't see this to be true if e.g. the vertex is the center of a disk embedding of the star $K_{1,\mathfrak{c}}$. The construction wouldn't change anything and not create free space using the Cantor fan.
• Concerning the edge, I just don't understand what "Just grab an arc and peel the space open a bit along it." is supposed to mean.

I would like you to prove that any graph with a spherical (combinatorical) embedding also has one on $\mathbb{R}^2$ (or provide a counterexample), either completely by yourself or expanding/explaining John's answer.

Definition of embedding to be used: A multigraph $G=(V,E,S,T)$ can be embedded (in the sense of this question) on a surface $S$ if there are injective functions $\phi:V\rightarrow S$, $\psi:E\rightarrow C([0,1],S)$ (the curves on $S$) with the following properties:

• $\forall e\in E:\phi(S(e))=(\psi(e))(0)$ and $\phi(T(e))=(\psi(e))(1)$, i.e. the endpoints of the curve representations of the edges are the corresponding representation of the vertices.
• $\forall e,f\in E, s,t\in[0,1]:e\neq f \wedge (\psi(e))(s) = (\psi(f))(t) \implies s,t\in\{0,1\}$, i.e. the presentations of two edges only intersect in vertices. This results in a planar drawing in case of $S=\mathbb{R}^2$.

In case of infinite graphs this definition can lead to embeddings which are not topological isomorphic to the topological space generated by the graph. This is explicitly allowed in this question.

Why this should not be treated as a duplicate of this question:

1. The other question wasn't clear on which kind of embedding to use, resulting in confusion. I'm very clear.
2. John's answer has many comments and several edits, not really helping my case, which is why I doubt further prodding would improve the answer's structure, a new one would be better.
3. If in doubt, my question is a proof verification of John's answer and I don't see why MSE answers couldn't be explained like statements in papers.