Show that for a continuous, increasing function $f : \mathbb R \rightarrow [0,1]$ that is onto $(0,1)$ it's inverse $f^{-1}$ is also continuous.
What I've tried
My first approach:
$f^{-1}$ is continuous if $f$ maps open sets to open sets. Since $f$ is increasing it is a one-to-one mapping and it suffices to show that $f$ maps open intervals to open intervals. Let $V \subset [0,1]$ so that $f^{-1}(V)$ is open (since $f$ is continuous) and $f(f^{-1}(V))=V$ which is also open. All that remains to show is that every open interval $U = f^{-1}(V)$ for some open interval $V$. This is true since $f$ is a one-to-one mapping.
My second approach:
$f^{-1}$ is continuous if there exists a $\delta$ such that when:
$$|\alpha - \alpha'| < \delta$$
we have:
$$|f^{-1}(\alpha) - f^{-1}(\alpha')| < \epsilon$$
Call the first event $A$ and the second event $B$. To show $A \implies B$ it is sufficient to show that $B^c \implies A^c$. Let $\alpha = f(x),\, \alpha' = f(x')$ for some $x,x'$ so that we need to show for each $\epsilon$: $|x-x'| \geq \epsilon \implies |f(x) - f(x')| \geq \delta$ for some $\delta$. But this is clear since $f$ is strictly increasing.
I could use some feedback on my reasoning for both approaches.