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 ultrametrizable 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.