To elaborate a bit, I should say that the question of the existence of a complete metric is only of interest in the case of manifolds of infinite topological type; if a manifold is compact, any metric is complete, and if a noncompact manifold has finite topological type(ie is diffeomorphic to the interior of a compact manifold with boundary,) one can contruct a complete metric on the manifold with boundary via a partition of unity, and then divide by the square of a defining function to get an complete asymptotically metric on the interior.


I have absolutely no intuition for how "wild" these manifolds can be. The only examples I can think of are infinite connected sums and quotients of negatively curved symmetric spaces by sufficiently complicated groups, but I'd imagine that one can construct some pathological examples by limiting arguments.



Does every smooth manifold of infinite topological type admit a complete Riemannian metric?
By Whitney embedding theorem any smooth manifold embeds into some Euclidean space as a closed subset. The induced metric is complete.


In fact, a good exercise is to show that any Riemannian metric is conformal to a complete metric.


