Are these equivalent characterizations of closed manifolds? Let $M$ be a connected smooth manifold without boundary. Are the following equivalent?


*

*$M$ is compact

*$M$ cannot be realized as a proper open subset $M\subset N$ of another connected manifold $N$. In other words, $M$ cannot be made bigger without adding connected components.

*The flow of every vector field on $M$ is complete.


I recall wondering about this as a student. Some of the implications are well know but I never figured out the others. This question reminded of this.
 A: That 1 implies 3 is standard and appears in most textbooks on manifolds that include vector field flows, it follows by a standard existence and uniqueness argument. 
3 implies 1 is a little more work but it's a fairly standard argument as well.  The idea is that if the manifold is non-compact you can take a proper Morse function on the manifold, $f : M \to [0,\infty)$.   If you re-scale the gradient like $\frac{f^2}{1+|\nabla f|^2}\nabla f$, its flow lines can't be complete since it hits infinity in a finite time.   And technically you don't need any Morse theory to make this conclusion.  It's enough to have a proper function $f : M \to [0,\infty)$ then show you can embed an arc $p : [0,\infty) \to M$ in $M$ so that $f(p(x)) = x$ for all $x$.  You can define an incomplete vector field whose flow line is the image of $p$, then extend it to $M$. 
To relate 1 and 2 notice that in a manifold, a subspace is compact, connected and open means its a path-component.  If you take a manifold and collapse each path-component to a point (the path-component space) you get a discrete space.  So this gives you 1 implies 2.  The converse 2 implies 1 follows from the argument in my comment to Mariano above -- if it's non-compact you can take a properly embedded arc, cut along it and self-embed.  In reverse this embeds your manifold in a bigger manifold. 
