Let $f$ be strictly increasing on $I$. Prove that $f$ has an inverse defined on its range and $f^{-1}$ is continuous.
The existence of the inverse is simple (if $y = f(x)$, set $x = f^{-1}(y)$, since $f$ is strictly increasing).
However, I'm difficult understanding the proof for the fact that the inverse is continuous. I have looked it up on the internet and I don't follow their choices of $\delta$ and what they're doing. If someone could explain the proof, that'd be brilliant.