Prove that $f:(a, b) \to \mathbb R$ has at most countably many simple discontinuities

Let $E_n = \left\{ x \in \left( a , b \right) \mid \lvert \lim_{t \to x^+} f(t) - \lim_{t \to x^-} f(t) \rvert > \frac{1}{n} \right\}$, the set of simple discontinuities whose jumps are greater than $\frac{1}{n}$. Then $E_1 \subset E_2 \subset\cdots$ and $\cup^\infty_{i=1} E_i$ is the set of all simple discontinuities. If all of the $E_n$ are countable, then $\cup^\infty_{i=1}E_i$ is countable.

Suppose $ x \in E_n $.

The left-hand limit exists, so

$$ \exists \delta^-_x > 0 \colon t \in \left( x - \delta^-_x, x \right) \implies \lvert f(t) - \lim_{t \to x^-} f(t) \rvert < \frac{1}{4n} $$

Suppose $y \in \left( x - \delta^-_x, x \right)$ is a simple discontinuity. Then

$$ \exists \delta^-_y > 0 \colon t \in ( y - \delta^-_y, y ) \implies \lvert f(t) - \lim_{t \to y^-} f(t) \rvert < \frac{1}{4n} \\ \exists \delta^+_y > 0 \colon t \in ( y, y + \delta^+_y ) \implies \lvert f(t) - \lim_{t \to y^+} f(t) \rvert < \frac{1}{4n} $$

Then $\exists z^- \in (y - \delta^-_y, y) \cap (x - \delta^-_x, x), z^+ \in (y, y + \delta^+_y) \cap (x - \delta^-_x, x) \colon$

$$ \lvert \lim_{t \to y^+} f(t) - \lim_{t \to y^-} f(t) \rvert \leq \\ \lvert \lim_{t \to y^+} f(t) - \lim_{t \to x^-} f(t) \rvert + \lvert \lim_{t \to y^-} f(t) - \lim_{t \to x^-} f(t) \rvert \leq \\ \lvert \lim_{t \to y^+} f(t) - f(z^+) \rvert + \lvert \lim_{t \to x^-} f(t) - f(z^+) \rvert + \lvert \lim_{t \to y^-} f(t) - f(z^-) \rvert + \lvert \lim_{t \to x^-} f(t) - f(z^-) \rvert \leq \\ \frac{1}{n} $$

So that

$$ \forall x \in E_n \exists \delta > 0 \colon y \in (x - \delta, x ) \implies y \not \in E_n $$

This says that for a jump at $x$, there exists a left-hand neighborhood around x such that the jumps in the neighborhood are smaller than the jump at $x$ (we could prove the same thing for a right-hand neighborhood, but that turns out to be unnecessary). For each $x$, choose a rational number $q \in \mathbb{Q}$ from $(x - \delta, x )$, e.g.,

$$ \frac{ \lceil (x - \delta) ( \lceil \frac{1}{\delta} \rceil + 1 ) \rceil + 1}{\lceil \frac{1}{\delta} \rceil + 1} $$

Since the $(x - \delta, x)$ are disjoint, $g \colon x \mapsto q$ is an injection $g \colon E_n \rightarrowtail \mathbb{Q}$ (each $x$ maps to a different $q$). Since $\mathbb{Q}$ is countable, so is $E_n$.

so there must be intervals to the left and right of a simple continuity on which no other simple discontinuity can exist.

That is incorrect. Consider an enumeration $r_n$ of the rationals, and let

$$f(x) = \sum_{\substack{n \in \mathbb{N}\\r_n < x}} 2^{-n}.$$

Then $f$ is a strictly monotonic function that has a jump discontinuity in every rational number.

However, as the points of discontinuity approach any fixed $y \in \mathbb{R}$, the jumps tend to $0$, in fact, the sum of all jumps in the discontinuities $\neq y$ contained in a neighbourhood of $y$ tends to $0$ when the neighbourhood shrinks to a point.