So, here's the theorem that I'm trying to prove. The thing is, the book I'm using is somewhat older than other textbooks, so I'm trying to put into language that I'm more familiar with. I'm going to try and rewrite the statement of the theorem in my own words and then I'll try to prove it.
I'm hoping that someone can point towards any mistakes I'm making in my interpretation of the language being used. I'm only going to do it for the first portion, as the second part is going to be analogous.
Let $I \subset \mathbb{R}$ be an interval and let $f: I \to \mathbb{R}$ be a function defined on that interval. Suppose that $f$ is increasing. Then, the following hold:
$f$ is injective
$f^{-1}$, if it exists, is increasing.
Proof Attempt:
We have to show that:
$$\forall x,y \in I: x \neq y \implies f(x) \neq f(y)$$
Let $x \neq y$. Then, by the trichotomy law, $x > y$ or $y > x$. Since $f$ is increasing, we have either $f(x) > f(y)$ or $f(y) > f(x)$. In any case, we have $f(x) \neq f(y)$. That proves that $f$ is injective.
Now, if $f^{-1}$ exists, then $f$ is bijective and $f^{-1}$ is also bijective. Since $f$ is increasing, we have:
$x<y \implies f(x)<f(y)$
Let $b_1,b_2 \in f(I)$. By the bijectivity of $f$, there exist $a_1,a_2 \in I$ such that $f(a_1) = b_1$ and $f(a_2) = b_2$. Now, we have to show that:
$b_2 > b_1 \implies a_2 > a_1$
Let $a_1 \geq a_2$. Since, $f(a_1) = f(a_2) \iff a_1 = a_2$ and $a_1 > a_2 \implies f(a_1) > f(a_2)$, it must be the case that $a_1 \geq a_2 \implies f(a_1) \geq f(a_2)$. That is precisely the contrapositive of the statement we wanted to show, so we have proven that the inverse is increasing.
This proves the desired result.