If $F$ is a $1$-to-$1$ imersion and is proper then $F$ is an imbedding 
Let $F:N\rightarrow M$ be a one-to-one immersion which is proper, i.e. the inverse image of any compact set is compact. Show that $F$ is an imbedding and that its image is a closed regular submanifold of $M$ and conversely.

This is an exercise from Bothby: an introduction to diferentiable manifolds. Pag. 81, ex. 6
the following theorem say something very similar on what the exercise asks 
if $F:N\rightarrow M$ is an one-to-one immersion and $N$ is compact, the $F$ is and imbedding and $F(N)$ is a regular submanifold. Thus a submanifold of $M$, if compact in $M$, is regular.
Well, this theorem states exactaly what i want to show, but my dificulty is how changing the hypothesis of $N$ being compact for $F$ being proper could take me to the desired result?

Assume that it is not an embedding. For $p\in f(N)$, consider $$ f(N)\ \bigcap\ B\bigg(p,\frac{1}{n}\bigg)
 \supseteq U\cup \{p_n\}$$ where $U$ is homeomorphic to
 ${\rm dim}\ N$-dimensional open ball.
Hence $p_n\rightarrow p$ so that $\{p_1,\cdots \}\cup B(p,\varepsilon) $ is compact. Its preimage contains
  $U'$ and infinite points $p_n'\in f^{-1}(p_n)$ where $U'$ is homeomorphic to a ${\rm dim}\ N$-dimensional open ball.
By proper, $p_n'$ converges to $q$ which is not in $U'$. Then
$f(p_n')\rightarrow f(q)\neq p$. It is a contradiction. 
