Here is Prob. 17, Chap 4 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Let $f$ be a real function defined on $(a, b)$. Prove that the set of points at which $f$ has a simple discontinuity is at most countable. Hint: Let $E$ be the set on which $f(x-) < f(x+)$. With each point $x$ of $E$, associate a triple $(p, q, r)$ of rational numbers such that $$\begin{align}& (a)\ \ \ \ \ \ f(x-) < p < f(x+). \\ &(b) \ \ \ \ \ a < q < t < x \ \mbox{ implies } \ f(t) < p. \\ &(c) \ \ \ \ \ x < t < r < b \ \mbox{ implies } \ f(t) > p. \end{align} $$ The set of all such triples is countable. Show that each triple is associated with at most one point of $E$. Deal similarly with the other possible types of simple discontinuities.
Here is Definition 4.26 in Baby Rudin, 3rd edition:
Let $f$ be defined on $(a, b)$. If $f$ is discontinuous at a point $x$, and if $f(x+)$ and $f(x-)$ exist, then $f$ is said to have a discontinuity of the first kind, or a simple discontinuity, at $x$. Otherwise the discontinuity is said to be of the second kind.
My effort:
Let $x_1 < x_2$ be any two points of $E$, and let $\left( p_1, q_1, r_1 \right)$ and $\left( p_2, q_2, r_2 \right)$ be their respective triples of rational numbers, as given in the hint. If $p_1$ is not between $f\left( x_2-\right)$ and $f\left(x_2+\right)$, then $p_1 \neq p_2$ and so these two triples are different.
What if $f\left( x_2 - \right) < p_1 < f\left( x_2 + \right)$? How then to show that the two triples are still distinct?
Once we show this, then the set of all such triples is a subset of $\mathbb{Q}^3$ and is therefore countable.
Let $G$ be the set of all the points $x$ of $(a,b)$ for which $\lim_{t \to x} f(t)$ exists, but $f(x) \neq \lim_{t\to x} f(t)$. How to show that this set is also countable by associating a triple of rational numbers to each point of $G$?
Now let $F$ be the set of points $x$ in $(a,b)$ for which $f(x-) > f(x+)$. Then $F$ is countable. This we can show just like the proof for $E$. Right?
My attempt (after the hint by Doug M).
Let $x_1 < x_2$ be any two points of $E$, and let $\left( p_1, q_1, r_1 \right)$ and $\left( p_2, q_2, r_2 \right)$ be their respective triples of rationals, as given in the hint. Then we have $$ (a) \ \ f\left(x_1-\right) < p_1 < f\left( x_1+ \right) \ \mbox{ and } \ f\left( x_2-\right) < p_2 < f\left( x_2+ \right),$$ $$ (b) \ \ a < q_1 < t < x_1 \ \mbox{ implies } \ f(t) < p_1 \ \ \mbox{ and } \ \ a < q_2 < t < x_2 \ \mbox{ implies } \ f(t) < p_2,$$ $$ (c) \ \ x_1 < t
p_1 \ \ \mbox{ and } \ \ x_2 < t < r_2 < b \ \mbox{ implies } \ f(t)> p_2.$$ Let $c$ be a rational number such that $$ x_1 < c < x_2, $$ and then re-adjust our choice of the triples such that $$ r_1 < c < q_2 .$$ That is, while choosing $\left( p_1, q_1, r_1 \right)$, we focus on the smaller segment $(a, c)$ rather than on the entire segment $(a, b)$, and then while choosing $\left( p_2, q_2, r_2 \right)$ we focus on $(c, b)$ rather than on $(a, b)$. in this way, we see that the triples $\left( p_1, q_1, r_1 \right)$ and $\left( p_2, q_2, r_2 \right)$ are distinct. Thus it follows that $E$ is at most countable.
Similarly, the set $F$ of points $x$ at which $f(x-) > f(x+)$ is at most countable.
Let $G$ be the set of points $x$ at which $\lim_{t \to x} f(t)$ exists but is not equal to $f(x)$. Then how to show that this set $G$ is at most countable?
We break $G$ up into two sets $G_1$ and $G_2$ as follows: Let $G_1$ be the set of all the points $x$ of $(a, b)$ at which $\lim_{t \to x} f(t)$ exists and $f(x) < \lim_{t \to x} f(t)$, and let $G_2$ be the set of all those points $x$ of $(a, b)$ at which $\lim_{t \to x} f(t)$ exists and $ f(x) > \lim_{ t \to x} f(t)$. Then $G = G_1 \cup G_2$.
We now show that either of the sets $G_1$ and $G_2$ is at most countable.
Let $x$ be a point of $G_1$. Let $p$ be a rational number such that $$ f(x) < p < \lim_{ t \to x } f(t).$$ Now let us choose rational numbers $q$ and $r$ such that $$ a < q < t < x \ \mbox{ implies } \ f(t) > p,$$ and $$ x < t < r < b \ \mbox{ implies } \ f(t) > p.$$ Thus corresponding to every point $x$ of set $G_1$, we have a triple $(p, q, r)$ of rational numbers.
Now let $x_1 < x_2$ be any two points of $G_1$, and let $\left( p_1, q_1, r_1 \right)$ and $\left( p_2, q_2, r_2\right)$ denote their respective triples. Let $c$ be a rational number such that $x_1 < c < x_2$.
We first regard $f$ to be a function on $(a, c)$. Then $\left( p_1, q_1, r_1 \right)$ be a triple of rational numbers such that $$f\left( x_1 \right) < p_1 < \lim_{t \to x_1} f(t),$$ $$ a < q_1 < t < x_1 \ \mbox{ implies } \ f(t) > p_1,$$ and $$ x_1 < t < r_1 < c \ \mbox{ implies } \ f(t) > p_1.$$
Now we consider $f$ to be a function on $(c, b)$ rather than on $(a, b)$. Then we have a triple $\left( p_2, q_2, r_2 \right)$ of rational numbers such that $$ f\left( x_2 \right) < p_2 < \lim_{t \to x_2} f(t),$$ $$ c < q_2 < t < x_2 \ \mbox{ implies } \ f(t) > p_2,$$ and $$ x_2 < t < r_2 < b \ \mbox{ implies } \ f(t) > p_2.$$ Corresponding to any two points $x_1 < x_2$ in $G_1$, we can choose triples $\left( p_1, q_1, r_1 \right)$ and $\left( p_2, q_2, r_2 \right)$ of rational numbers such that there is a rational number $c$ such that $$q_1 < x_1 < r_1 < c < q_2 < x_2 < r_2,$$ which implies that $q_1 < q_2$ and $r_1 < r_2$ and hence that these two triples are distinct.
Thus the mapping $x \mapsto (p, q, r)$ of $G_1$ into $\mathbb{Q}^3$ is injective, which implies that $G_1$ is at most countable.
Similarly, the set $G_2$ is at most countable, from which it follows that the set $G$ is at most countable.
Is my reasoning correct? If so, then have I managed to cover all possible cases?