real analysis - A continuous bijection $f:mathbb{R}to mathbb{R}$ is an homeomorphism?

A continuous bijection $f:\mathbb{R}\to \mathbb{R}$ is an homeomorphism. With the usual metric structure.

I always heard that this fact is true, but anyone shows to me a proof, and I can't prove it. I was tried using the definition of continuity but it is impossible to me conclude that. I tried using the fact that $f$ bijective has an inverse $f^{-1}$ and the fact that $ff^{-1}$ and $f^{-1}f$ are continuous identity, but I can't follow.

Answer

I think that I have the answer using Michael Greinecker hint. I'll first prove it, because I did not saw this theorem before.

Hint: A continuous injective function from the reals to the reals must be strictly increasing or strictly decreasing.

Proof: Let $I=[a,b]$ be a closed interval with $af(y)$ by injectivity can't be the equality. If $f(a)

If image is all, I'll prove that $f^{-1}=g$ is continuous. Is well known that if $f$ is strictly increasing then $g$ so is. I'll prove continuity in some $a\in\mathbb{R}$. Let $\epsilon>0$ then $g(a)-\epsilon/2$ and $g(a)+\epsilon/2$ have preimages let $g(b)=g(a)-\epsilon/2$ and $g(c)=g(a)+\epsilon/2$ then $b