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.