If $X$ and $Y$ are completely metrizable spaces, let us write $X \preceq Y$ whenever $X$ embeds in $Y$.

>*Is this a well-quasi-order?*

Recall that $\preceq$ is a well-quasi-order if and only if:

$\ $ (1) there is no infinite $\preceq$-antichain (in this context, an infinite set of completely metrizable spaces, none of which embeds in another).

$\ $ (2) there is no infinite strictly decreasing sequence (a sequence $X_0, X_1, X_2, \dots$ of completely metrizable spaces such that $X_{n+1} \preceq X_n$ but $X_n \not\preceq X_{n+1}$ for all $n$).
    
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**Further comments:** 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][1]), together with the fact that every completely ultrametrizable space can be represented as a tree.

This all seems very suggestive to me -- hence the question.

Lastly, I know that beggars can't be choosers, but I would be especially interested to see a counterexample involving only Polish spaces.


  [1]: http://web.mat.bham.ac.uk/D.Kuehn/bqofinal.pdf