This is Problem 3 in Guillemin & Pallock's Differential Topology on Page 18. So that means I just started and am struggling with the beginning. So I would be expecting a less involved proof:
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a local diffeomorphism. Prove that the image of $f$ is an open interval and maps $\mathbb{R}$ diffeomorphically onto this interval.
I am rather confused with this question. So just identity works as $f$ right?
The derivative of identity is still identity, it is non-singular at any point. So it is a local diffeomorphism.
$I$ maps $\mathbb{R} \rightarrow \mathbb{R}$, and the latter is open.
$I$ is smooth and bijective, its inverse $I$ is also smooth. Hence it maps diffeomorphically.
Thanks for @Zev Chonoles's comment. Now I realized what I am asked to prove, though still at lost on how.
Hint: Since $f$ is a local diffeomorphism, $f$ is a local homeomorphism. Use this fact to prove that $f$ is an open map. By continuity, the image of $f$ is connected. What is the only type of open, connected subset of $\mathbb{R}$?
Here's the proof that $f$ is an open map. Let $U \subseteq \mathbb{R}$ be an open set. Given $y \in f(U)$, there exists $x \in U$ such that $f(x) = y$. Take an open set $U_x$ around $x$ for which $f_{U_x}$ is a homeomorphism onto its image. Since $U \cap U_x$ is an open subset of $U_x$, $f(U \cap U_x)$ is open in $f(U_x)$. However, $f(U_x) \subseteq \mathbb{R}$ is open, and therefore, $f(U \cap U_x) \subseteq \mathbb{R}$ is open. We have
$$y \in f(U \cap U_x) \subseteq f(U)$$
and so $f(U)$ is open.
For the second part, it suffices to show that the local diffeomorphism $f: \mathbb{R} \to (a, b)$ is a bijection. We know it's surjective, so assume it's not $1-1$. Then by Rolle's theorem (or MVT), there is some point $x \in \mathbb{R}$ such that $f'(x) = 0$. However, the pushforward $df_x$ at $x$ must be a linear isomorphism. This is a contradiction.