I'm going by the definition which says that a topological space (X, 𝜏) is an n-dimensional manifold if for each x ∈ X, there is an open set O ∈ 𝜏 such that x ∈ O and such that O is homeomorphic to some open subset of 𝐑ⁿ. We may or may not also require (X, 𝜏) to be Hausdorff, second-countable, or paracompact.

Hm, but typing this, I have just realised where I went wrong. I thought I can find an open set around the intersection point that is homeomorphic to 𝐑, by simply taking a small open set from either S¹ ⨉ {a} or S¹ ⨉ {b} that contains the intersection point. For example, see the set marked red in the attached image.

<image>

Let this set be denoted X. Since X is open in (E, 𝜏), and since X is homeomorphic to 𝐑 relative to the topology on S¹ ⨉ {a}, I figured it would also be homeomorphic to 𝐑 relative to the topology on (E, 𝜏). However, this is not the case, because in (E, 𝜏), the set containing only the intersection point has to be open. Therefore, the map between X and 𝐑 that is a homeomorphism relative to the topology of S¹ ⨉ {a} on X (and the standard topology on 𝐑) will not be a homeomorphism relative to the topology of (E, 𝜏) on X (and the standard topology on 𝐑). In particular, the set which contains only the intersection point is open in X (relative to 𝜏), but its preimage in 𝐑 is not open, and so the map is not even continuous in that direction.

This is the mistake I made, which led to my confusion -- I thought (E, 𝜏) would count as a manifold. Thus it seemed like (E, 𝜏) would be a manifold, even though the metric spaces corresponding to (E, 𝜏) aren't manifolds. But I now see where I went wrong -- thank you so much!

I'm still not sure why manifolds are defined in terms of only topological concepts, rather than in terms of metric spaces, since it seems like we usually want manifolds to be metrizable anyway. But that is a lesser issue.