My question reads: Prove that if the real-valued function $f$ is increasing (or decreasing), then f is one-to-one.
Here is my proof:
Let $f:\Bbb{R}\to\Bbb{R}$ be increasing (or decreasing) on $\Bbb{R}$
Let $x_1,x_2 \in \Bbb{R}$ s.t. $x_1<x_2$
Case 1
$f$ is increasing then $f(x_1)< f(x_2)$
$x_1< x_2$, so $x_1$ doesn't equal $x_2$, also $f(x_1)< f(x_2)$, then $f(x_1)$ doesn't equal
$f(x_2)$ since $x_1$ doesn't equal $x_2$. Thus $f$ is one to one.
Case 2
$f$ is decreasing then $f(x_1) >f(x_2)$
$x_1< x_2$, so $x_1$ doesn't equal $x_2$, also $f(x_1) >f(x_2)$, then $f(x_1)$ doesn't equal $f(x_2)$ since $x_1$ doesn't equal $x_2$. Thus f is one to one.
You start by saying that you assume $x_1 < x_2$. What follows is basically correct under that assumption. You did not, however, prove the result in the case that $x_1 > x_2$.
You could introduce cases 3 and 4 that prove things in the case $x_1 > x_2$ with $f$ increasing, and $x_1 > x_2$ with $f$ decreasing. However, the way I would approach this would be to replace the statement "Let $x_1, x_2\in\mathbb{R}$ with $x_1 < x_2$" with the following:
"Assume $x_1\ne x_2$. Then either $x_1 < x_2$ or $x_1 > x_2$. If $x_1 < x_2$, reverse the roles of $x_1$ and $x_2$. So in either case we may assume that $x_1 < x_2$."
Then proceed exactly as you did.