During this question a *manifold* $M$ is meant to be a smooth **connected** (second countable, finite dimensional) manifold.


A Riemannian metric on a manifold $M$ is called *complete* if every geodesic is defined for all times. It is well-known that this is equivalent to completeness of the metric space $(M,d)$ where $d$ is the geodesic distance.


I want to show the following equivalence:
A manifold $M$ is compact if and only if every Riemannian metric on $M$ is complete. (keep in mind that $M$ is connected)


It is easy to show the implication from left to right, but I have no clue for the reverse implication.


I tried the following argument:
Every smooth manifold $M$ can be embedded as a bounded submanifold of $\mathbf R^n$. Assume that $M$ is not compact, then $M$ is not closed in $\mathbf R^n$. So, as a metric space $M$ is not complete when given the metric of the surrounding space $\mathbf R^n$. 
Then I realized that this does not imply that $M$ with the induced Riemannian metric is not complete since the geodesic distance on $M$ may be totally different from the induced distance function.


So my question is: If $M$ is non-compact, does there exist an incomplete Riemannian metric on $M$?



Existence of incomplete Riemannian metrics
If $M$ is non-compact, then there exists a sequence of points with no convergent subsequence. Fix a curve passing through all of the points. It suffices to construct a Riemannian metric such that this curve has finite length. 


