TITLE: Every continuous open mapping $\mathbb{R} \to \mathbb{R}$ is monotonic
QUESTION [1 upvotes]: Prove that every continuous open mapping from $\mathbb{R} \to \mathbb{R}$ is monotonic

I want to prove it only (or mostly) using arguments and concepts from topology, and not from analysis.
I don't have anything that I think is useful or correct yet. The only idea I had was this:
Let $f:\mathbb{R}\to\mathbb{R}$ be continuous and open. If $x,y \in \mathbb{R}$, $x<y$, then $(x,y)$ is connected and open. Since $f$ is open, $f(x,y)$ must be open, and since $f$ is continuous, $f(x,y)$ must be connected. This way, if I managed to prove that $f(x,y)$ is bounded, then we would have $f(x,y)=(a,b)$ for some $a,b\in\mathbb{R}$. Now, because $f$ is continuous and open, we would have that $f(\partial(x,y))=\partial(a,b)$, that is, $\{f(x),f(y)\}=\{a,b\}$, which implies that $f(x)=a, f(y)=b$, or $f(x)=b, f(y)=a$ and finally $f(x)<f(y)$ or $f(x)>f(y)$.
However, I don't know how to prove that $f(x,y)$ is bounded, and even if I did it, I think this would only establish that f is injective, not that it is monotone.
I would be thankful if you could give me some hints (not full solutions).

REPLY [4 votes]: Hint: Supose that $f$ is not monotonic, then exist a interval $[x,y]$ and a point $t \in (x,y)$, such that $f(t)= \max_{s \in [x,y]}{f(s)}$ or $f(t)= \min_{s \in [x,y]}{f(s)}$. Once you have that, study the set $f((x,y))$.