Question: is it true that every finite-dimensional connected complete Riemannian manifold of non-positive sectional curvature is a Polish space, i.e. homeomorphic to a complete separable metric space?

My reasoning is that by the Cartan-Hadamard theorem, a universal covering space of a connected complete Riemannian manifold of non-positive sectional curvature is diffeomorphic, and thus also homeomorphic, to a Euclidean space. The space itself is a universal covering space by taking the identity map as a universal cover, and the Euclidean space is a complete separable metric space.

anyconnected smooth manifold $M$ can beproperlyimbedded in an Euclidean space $E$. In other words you can view $M$ as aclosedsubset of an Euclidean space with the induced topology. As such it is a Polish space. For details on Whitney's imbedding see Chapter IV Sect 1 of Whitney'sGeometric Integration Theory. $\endgroup$ – Liviu Nicolaescu Mar 3 '17 at 11:14