Does any manifold admit a complete distance metric? 
Let $M$ be a smooth manifold. Is there a metric $d$ on $M$ compatible with the original topology such that $(M,d)$ is a complete metric space? Note that in this question I always mean distance metric, not a Riemannian metric.

My initial idea was to start with a cover on $M$ with charts $U_i$ diffeomorphic to $\mathbb{R}^n$  and then pullback the Euclidian distance. Denote it by $d_i$. Glue everything with a partition of unity $(a_i)$ subordinated to the cover $U_i$, in the sense that consider $d=\Sigma a_id_i $. But I don't know if it is a correct approach. I also don't know if the answer to my question is affirmative.

The Hopf-Rinow Theorem states that a (connected) Riemannian manifold $(M, g)$ is complete if and only if the metric $d$ on $M$ induced by $g$ is complete. (Here, $d$ is the Riemannian distance determined by $g$: $$d(x, y) := \inf_\gamma \int_\gamma ds = \int_{t_0}^{t_1} g(\gamma'(t), \gamma'(t)) \,dt ,$$ where $\gamma$ varies over the paths from $x$ to $y$.) So, it suffices to show any manifold admits a complete Riemannian metric $g$.
In fact, something much stronger is true: For any Riemannian manifold $(M, g)$ there is some metric conformal to $g$---that is, a metric of the form $\lambda g$ where $\lambda$ is a smooth, positive function---that is geodesically complete (see this paper of Nomizu and Ozeki [pdf warning]). Thus, it suffices to show that any smooth manifold admits at least one Riemannian metric, but this is a standard exercise using partitions of unity.
