Every non empty paracompact connected 1-manifold is either homeomorphic to a circle or to the real line. Therefore, one can trivially say that all 1-manifolds (without boundary) covered by a single chart are equivalent to $\mathbb{R}$.
Is there a similar result for higher-dimensional manifolds? At the moment, I cannot think of any 2-manifold entirely covered by a chart and non-equivalent to $\mathbb{R^2}$. Am I wrong?
If in your definition a chart map is a homeomorphism from an open subset of $M$ to $\mathbb{R}^n$, and you have a single chart that cover your whole manifold, then $M$ is obviously homeomorphic to $\mathbb{R}^n$.
If in your definition a chart map is a homeomorphism from an open subset of $M$ to an open (connected) subset of $\mathbb{R}^n$, the the open annulus is an open 2-manifold that allow a single chart but that is not $\mathbb{R}^{2}$.
It depends on your definition of "chart" (even in dimension one).
If you merely require the domain of a chart to be an open subset of $\mathbf{R}^{n}$, then every open subset (e.g., a disjoint union of intervals, an annulus...) is a manifold covered by a single chart, and "most" of these are not diffeomorphic to $\mathbf{R}^{n}$.
If your definition requires a chart domain to be diffeomorphic to $\mathbf{R}^{n}$, then your intuition is of course correct. ;)
