Spaces that can't be embedded in the plane If $X$ and $Y$ are topological spaces, let us write $X \preceq Y$ whenever $X$ embeds in $Y$.
Earlier today, I asked the question:

Is this a well-quasi-order on the completely metrizable spaces?

This was short-sighted, as Tom Goodwillie has pointed out in the comments that the closed surfaces give an easy counterexample.
Since I can't accept Tom's comment as an answer, I'd like to modify the question to make it more interesting (while still being very closely related to the original):

Is there a finite list $F$ of completely metrizable spaces such that, for any completely metrizable space $X$, $X \preceq \mathbb{R}^2$ if and only if $Y \not\preceq X$ for every $Y \in F$?

An affirmative answer would be something analogous to Wagner's Theorem, but with a more topological flavor.
[Considering this question was part of what led me to ask my other question: if embeddability were a wqo (which it isn't), then the answer to the present question would be yes.]
Candidate list: the topological graphs $K_5$ and $K_{3,3}$, the sphere $S^2$, and the subspace of $\mathbb{R}^3$ obtained by taking the X-Y plane and a sequence converging to the origin along the Z axis.
[Notice that every closed surface contains one of these.]

The following was a comment to the original question. It is not relevant to the modified question, but I am keeping it to explain the post of Nash-Williams below:
Embeddability is not a well-quasi-order for metric spaces generally. An easy way to get a counterexample is to build one by transfinite recursion: you can find infinitely many subsets of $\mathbb R$ that violate either/both of the conditions listed above. The examples you build will be very far from $G_\delta$, so not completely metrizable.
Completely ultrametrizable spaces are well-quasi-ordered by embedability. This follows (with a little bit of work) from a version of the Nash-Williams Tree Theorem (see Theorem 11 here), together with the fact that every completely ultrametrizable space can be represented as a tree.
 A: It seems that the answer to the question is negative. However, it should be noted that I did not work out all details for the following counterexample.
Consider a graph $G$ that is not planar. Replace every vertex of $G$ by a segment and every edge of $G$ by a double-sided topologist's sine curve, obtaining a space $X_G$. Then


*

*$X_G$ does not seem embeddable into $\mathbb R^2$.

*If we only consider $G$'s which are subdivisions of $K_5$, then they do not embed one into another. And it also seems that they cannot be forbidden by a finite list of non-embeddable spaces.


Both claims would require some bit of work to be proved properly.
A: You get a positive answer if you restrict to compact, locally connected metric spaces.
See
https://www.semanticscholar.org/paper/On-planarity-of-compact%2C-locally-connected%2C-metric-Richter-Rooney/70090a524f2509408e5170f814ed1b33654bb585
and references therein.
A: Edit: This is not an answer to the question, but a pointer to the solution of a different problem (due to a misunderstanding on my part)
I'm not an expert on this, so possibly misunderstood something, but to my mind
your result on ultrametric spaces appears to badly contradict Theorem 4.2 of a paper by Louveau and Rosendal, which says that the quasiordering of embeddability for Polish ultrametric spaces is a universal analytic quasi-order (for Borel reducibility), which implies that it is very far from being wqo.
The paper in question is available here: http://homepages.math.uic.edu/~rosendal/PapersWebsite/CompleteAnalytic.pdf
