Understanding an argument for a countable number of discontinuities of a monotone function I came across this argument but I'm having some trouble with it. $f$ is from $\mathbb{R} \to \mathbb{R}$, $f(a\pm 0)$ is the right and left handed limits of $f$ at $a$ respectively.

Assume that $f$ is monotone increasing on an interval $I$.
If $f$ is not continuous at a $c \in I$, then $f(c-0)\lt f(c+0)$. Let $r(c)$ be a rational number for which $f(c-0)\lt r(c) \lt f(c+0)$. If $c_1 \lt c_2$, then by the monotonicity of $f$, $f(c_1 + 0) \le f(c_2 - 0)$. Thus if $f$ has both $c_1$ and $c_2$ as points of discontinuity, then $r(c_1) \lt r(c_2)$. This means we have crated a one-to-one correspondence between the points of discontinuity and a subset of the rational numbers. Since the set of rational numbers is countable, $f$ can only have a countable number of discontinuities

However, I cant see what's wrong with saying, that since the irrationals are everywhere dense, you choose an irrational number $s$ such that $f(c-0)\lt s(c) \lt f(c+0)$, and conclude that you have a one-to-one correspondence of the discontinuities and a subset of the irrational numbers, concluding that there can be an uncountable number of discontinuities, which makes me think I don't fully understand the argument.
 A: This is a standard case of arriving, at some time during solving a problem, at a situation where you can make many different choices to proceed. You could chose a rational number from that interval, an irrational number, a transcendental number or many other things.
However, for the goal you want achieve by solving the problem, some choices are better than others.
To give an example, you leave a meeting at 1PM and need to be at the train station at 2PM, to catch your train. Depending on how far away the train station is, how much luggage you have, making a walk on foot may be perfectly reasonable, so you don't waste your money on a taxi (cab). Or taking public transport is feasable. Or maybe your only option is to really take a taxi and promise the driver a bonus if they reach it before 1:55.
All of those options may be available, but for your specific purpose, many could work, or maybe only one or even none.
It's the same with your proof. The goal of the proof is to arrive at an "as low as we can get" upper bound on the number of discontinuities of such functions. Since each discontiuity is a real number, knowing there are "at most as many discuntinuities as there are real numbers" is the starting point.
If you know a bit about irrational and transcendental numbers, you know they could also have been picked from in the proof. However, the result those choices would give is just not any better than what we knew beforehand. Of course to verify that you need to know that the cardinality of the irrational and transcendental numbers is the same as the cardinality of the real numbers.
That explains why choosing the rational numbers instead of the other mentioned is better for this specific purpose, because their cardinality is actually less than those of the real numbers.
A: A subset of the set of all irrational numbers may be countable, so a set which is in bijection with such a set can be countable.
