Lemma. Assume that $I$ is an open interval and $f:I\to\mathbb{R}$ is an injective, continuous map. Then $f(I)\subseteq\mathbb{R}$ is an open subset.

Proof. Since $f$ is injective, then of course $f$ is monotonic. Without loss of generality, we may assume that $f$ is increasing. Let $y=f(x)\in f(I)$ . Since $I$ is open, then there are $\alpha,\beta\in I$ such that $\alpha<x<\beta$ . Therefore $f(\alpha)<y<f(\beta)$ and (because continuous functions are Darboux functions) for any $y'\in \big(f(\alpha),f(\beta)\big)$ there exists $x'\in I$ such that $f(x')=y'$ . This shows that $\big(f(\alpha),f(\beta)\big)$ is an open neighbourhood of $y$ contained in $f(I)$ and therefore (since $y$ was arbitrary) $f(I)$ is open. $\square$

Proposition. Assume that $I$ is an open interval and $f:I\to\mathbb{R}$ is an injective, continuous map. Then $f$ is a homeomorphism onto image.

Proof. Of course, it is enough to show that $f$ is an open map. But if $U\subseteq I$ is open, then there are disjoint, open intervals $I_{\alpha}$ such that $$U=\bigcup_{\alpha} I_{\alpha}.$$ Therefore we obtain continuous, injective maps $f_{\alpha}:I_{\alpha}\to\mathbb{R}$ which are restrictions of $f$ to $I_{\alpha}$ . By lemma we have that $f_{\alpha}(I_{\alpha})$ is open and therefore $$f(U)=f\bigg(\bigcup_{\alpha}I_{\alpha}\bigg)=\bigcup_{\alpha}f(I_{\alpha})=\bigcup_{\alpha}f_{\alpha}(I_{\alpha})$$ is open. This shows that $f$ is a homeomorphism onto image. $\square$