Kuratowski counterexample [duplicate]

Say that a set $X⊆\mathbb{R}^2$ is discrete if every point $x\in\mathbb R^2$ has a neighborhood $U$ so that $|U∩X|=1$. A planar drawing of an infinite graph is a drawing of $G$ where the vertices of $G$ form a discrete set and edges are represented by non-crossing polygonal arcs. Show that there is an infinite graph on acountable vertex set that cannot be drawn in the plane but has no subdivision of $K_{3,3}$ or $K_5$. (That is Kuratowski’s theorem fails for infinite graphs)

$G$ is said to be planar if it has a planar drawing as described above (in particular separability).

My idea is this. I'm pretty sure it's going to work, but I don't know how to prove it. Essentially swapping two vertices always makes the graph worse.

Solution 1:

The easiest counterexample I can think of is to take $K_4$ and stick an infinite chain on each vertex. No matter the planar embedding of $K_4$, one vertex is inside a topological circle, so the connected infinite chain is inside this same bounded set, so the embedding of the vertex set of this chain has a limit point, hence it's not a closed discrete subspace of the plane.

Solution 2:

I believe this will do.

Take any square. No matter how you position it on the plane, it must be finite, and each infinite end can be either inside or outside it, and they can't be both outside.

On a side note, using a well-known term in a subtly different meaning is generally a bad idea, and the term "planar" in graph theory is used, er, well... more than a little.