How to prove that the inverse of a continuous strictly monotone increasing function is continuous? (Terence Tao Analysis 1, Proposition 9.8.3) I'm having some problems proving the inverse is continuous. The hint in the book is to use the standard epsilon-delta definition of continuity. I believe the easiest route is a proof by contradiction, but with all of the quantifiers in the statement, I may be incorrectly negating the statement I am trying to prove. Also, I have at my disposal the intermediate value theorem, which most of my proof relies on. Below is the proposition:
Let $a < b$ be real numbers, and let $ f:[a, b] \to \mathbb{R} $ be a function which is both continuous and strictly monotone increasing. Then $f$ is a bijection from $[a, b]$ to $[f(a), f(b)]$, and the inverse $f^{-1}: [f(a), f(b)] \to [a, b]$ is also continuous and strictly monotone increasing.
Below is my attempt at a proof:
Let $x_1, x_2 \in [a, b]$ be real numbers such that $f(x_1) = f(x_2)$. From the trichotomy of the real numbers, we have that exactly one of the following is true: $x_1 = x_2$, $x_1 < x_2$, or $x_1 > x_2$. Suppose $x_1 \not = x_2$. Then, by definition of strictly increasing monotone functions, we have that $f(x_1) \not = f(x_2)$. Thus, $x_1 = x_2$, and $f$ is injective.
Now let $y \in [f(a), f(b)]$ be a real number. Then, by the intermediate value theorem, there exists a real number $c \in [a, b]$ such that $f(c) = y$. Thus, $f$ is a surjection from $[a, b]$ to $[f(a), f(b)]$. Since $f$ is both injective and surjective, we can conclude that $f$ is a bijection from $[a, b]$ to $[f(a), f(b)]$.
To show that $f^{-1}$ is strictly monotone increasing, let $y_1, y_2 \in [f(a), f(b)]$ be real numbers such that $y_1 < y_2$. Then,by the intermediate value theorem, there exist $x_1, x_2 \in [a, b]$ such that $f(x_1) = y_1$ and $f(x_2) = y_2$. Since $f$ is strictly monotone increasing, we have $x_1 < x_2$. Using the definition of an inverse, we have\begin{align*}f^{-1}(y_1) &= f^{-1}(f(x_1)) \\&= x_1 \\&< x_2 \\&= f^{-1}(f(x_2)) \\&=f^{-1}(y_2) \text{,}\end{align*}showing that $f^{-1}$ is strictly monotone increasing.
Finally, we will show that $f^{-1}$ is continuous. Let $y_0 \in [f(a), f(b)]$ be a real number, and let $\epsilon > 0 $ be a real number. As before, there exists a real number $x_0 \in [a, b]$ such that $f(x_0) = y_0$. Likewise, for any real number $y \in [f(a), f(b)]$, the intermediate value theorem tells us that there exists a real number $x \in [a, b]$ such that $f(x) = y$. We want to show that there exists a $\delta > 0 $ such that $ | f^{-1}(y) -f^{-1}( y_0) | < \epsilon$ for all $y \in [f(a), f(b)]$ such that $|y - y_0| < \delta$. This is equivalent to showing that there exists a $\delta > 0 $ such that $ | x - x_0 | < \epsilon$ for all $f(x) \in [f(a), f(b)]$ such that $|f(x) - f(x_0)| < \delta$. Written in the order we are more accustomed to, this is equivalent to showing that there exists a $\delta > 0 $ such that $|f(x) - f(x_0)| < \delta$ for all $x \in [a, b]$ such that $|x - x_0| < \epsilon$.
Suppose, for the sake of contradiction, that $f^{-1}$ is not continuous. That is, suppose for all $\delta > 0$, there exists an $\epsilon > 0$ such that $|f(x) - f(x_0)| \ge \delta$ for all $x \in [a, b]$ such that $|x - x_0| < \epsilon$.
I am not really sure where to go from here, and I'm not certain I correctly negated the statement that the inverse of $f$ is continuous. Any help is greatly appreciated.
P.S. This is not for any homework, just self study. I've never taken a class in analysis, so please feel free to point out anything I am doing wrong (or that is less than rigorous).
 A: $\newcommand{\ep}{\epsilon}$
$\newcommand{\de}{\delta}$
$\newcommand{\f}{f^{-1}}$
$\newcommand{\ga}{\gamma}$
Use this formulation of the problem:

For any $y_0\in (f(a),f(b))$ and any $\ep>0$, $\exists \de>0$ s.t. $\f(y_0-\de,y_0+\de)\subset (\f(y_0)-\ep,\f(y_0)+\ep)$

(the case where $y_0=f(a)$ or $f(b)$ is similar to the following, and just require you ignore either the left or right half of the intervals involved)
Set $\ga=\textrm{min}(\ep,\f(y_0)-(a),b-\f(y_0))$. Note here that $\ga\leq \ep$. It is easy to see that the set $(\f(y_0)-\ga,\f(y_0)+\ga)$ lies in $[a,b]$.
Now consider $(f(\f(y_0)-\ga),f(\f(y_0)+\ga))$. Because $\f$ is strictly increasing, it is easy to see that this interval is mapped under $\f$ into $(\f(y_0)-\ga,\f(y_0)+\ga)$.
Finally, simply set $\de=\min(f(\f(y_0)+\ga)-y_0,y_0-f(\f(y_0)-\ga)]$. The set $(y_0-\de,y_0+\de)$ is a subset of $(f(\f(y_0)-\ga),f(\f(y_0)+\ga))$, and so is sent into $(\f(y_0)-\ga,\f(y_0)+\ga)$ by $\f$.
Because $\ga\leq \ep$, we then have that $\f(y_0-\de,y_0+\de)\subset(\f(y_0)-\ep,\f(y_0)+\ep)$, as desired.
