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This is an attempt to rephrase the question so that it makes more sense. As mentioned in other answers, there is absolutely no hope for a homeomorphism classification of open subsets of $\mathbb R^n$ if $n>2$.

I think, a better question would be "what can be said about manifolds that are homeomorphic to open subsets of $R^n$, or equivalently, manifolds that can be covered by a single coordinate chart?".

Any such manifold is parallelizable i.e. its tangent bundle is trivial, but I do not know other obstructions. One could consider a parallelizable manifold (e.g. a Lie group) that is not homeomorphic to $R^n$, take any of its open subsets $U$ and then how does one decide whether $U$ can be covered by a single coordinate chart?

So a test question: remove finitely many points from your favorite compact Lie group, and try to see if it embeds in the Euclideas space of the same dimension. Surely, this kind of questions must be answered if one hopes for a classification. I do not know the answer.

Amusingly, any infinite dimensional separable Hilbert manifold is homeomorphic to a an open subset of a separable Hilbert space, i.e. $l_2$.

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This is an attempt to rephrase the question so that it makes more sense. As mentioned in other answers, there is absolutely no hope for a homeomorphism classification of open subsets of $\mathbb R^n$ if $n>2$.

I think, a better question would be "what can be said about manifolds that are homeomorphic to open subsets of $R^n$, or equivalently, manifolds that can be covered by a single coordinate chart?".

Any such manifold is parallelizable i.e. its tangent bundle is trivial, but I do not know other obstructions. One could consider a parallelizable manifold (e.g. a Lie group) that is not homeomorphic to $R^n$, take any of its open subsets $U$ and then how does one decide whether $U$ can be covered by a single coordinate chart?

Amusingly, any infinite dimensional separable Hilbert manifold is homeomorphic to a subset of a separable Hilbert space, i.e. $l_2$.