Show that if $f:[a,b] \to \mathbb{R}$ is continuous and increasing, then the inverse function $f^{-1}$ exists and is also continuous and increasing on the interval on which it is defined.
Proof
Show that $f$ is one-to-one
Since $f$ is strictly increasing on $[a,b]$, for each $x_n \in [a,b]$ with $n \in \mathbb{N}$, $f(x_n)<f(x_{n+1})$
Then if $f(x_n)<f(x_{n+1}), \Rightarrow x_n<x_{n+1}$
So when $f(x_n)\neq f(x_{n+1}), \Rightarrow x_n \neq x_{n+1}$. Thus, $f$ is one-to-one
Show that $f$ is onto
Take $y \in f([a,b])$
Since $f$ is continuous on $[a,b],$ we can find an $x \in [a,b]$ such that $f(x)=y$ for all $y \in f([a,b])$
Thus, $f$ is onto
Since $f$ is one-to-one and onto, $f$ is bijective, which means $f^{-1}$ exists
Now, since $f$ is defined on $[a,b]$, $f^{-1}$ is defined on the interval $[f(a),f(b)]$
We know that $(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}$ by the inverse function theorem which tells us that if $f'>0$ on an interval, then $(f^{-1})'>0$ on that interval.
Since $f$ is increasing on $[a,b], f'(x)>0$ on $[a,b]$. Then $(f^{-1}){'}(x)>0$ on $[a,b]$
Thus, $f^{-1}$ is increasing.
I can't figure out how to show that $f^{-1}$ is continuous based off of all this.
I feel like this proof is full of mistakes so feel free to show me where I'm going wrong. I'm having trouble seeing all this.
You have correctly established that $f^{-1}$ exists.
Now, use the definition of "increasing" to conclude that $f^{-1}$ is increasing.
From there, use the definition of continuity. Consider any $y \in (f(a),f(b))$. Let $\epsilon > 0$. We want to show that there is a $\delta > 0$ for which $$ |y - y'| < \delta \implies |f^{-1}(y) - f^{-1}(y')| < \epsilon. $$ Let $x = f^{-1}(y)$. Define $y_1 = f(x- \epsilon)$ and $y_2 = f(x + \epsilon)$. Let $\delta = \min\{y - y_1, y_2 - y\}$. We note that for any $y'$ with $|y - y'| < \delta$, we have $$ y_1 < y' < y_2 \implies f^{-1}(y_1) < f^{-1}(y') < f^{-1}(y_2) \\\implies f^{-1}(y) - \epsilon < f^{-1}(y') < f^{-1}(y) + \epsilon\\ \implies |f^{-1}(y') - f^{-1}(y)| < \epsilon, $$ which is what we wanted.
The proof where $y$ is equal to either endpoint is similar.