how to say every point of $D_n$ is an isolated point? Let $f$ be any real function. Show that $f$ has at most countably many jump discontinuities (discontinuities of the first kind).
The proof outline is Let $E$ be the domain of $f$.
For brevity, let $f_+(a)$ denote $\lim_{x→a+} f(x)$ and let $f_-(a)$ denote $\lim_{x→a-} f(x)$.
For each positive integer $n,$ define $D_n = \{a ∈ E : |f_+(a) − f_-(a)| ≥ 1/n\}$.
How to show that every point of $D_n$ is an isolated point. The hint is [Otherwise, a one-sided limit fails to exist.]
I don't know how to say every point of $D_n$ is an isolated point?
 A: I am writing a long answer. I am not proceedig in the way as has been hinted at. If we consider the hint, the answer might be much shorter.
Let us assume that $D_n$ is infinite. Otherwise every element should be isolated. Suppose an element $a$ of $D_n$ is not isolated. What does it mean? Given any $\delta>0$, the neighbourhood $(a-\delta, a+\delta)$ contains infinitely many points of $D_n$. Without loss of generality let us assume that given any $\delta>0$,  $(a-\delta, a) \cap E$ has infinitely many elements of $D_n$. Then it is possible to construct a sequence $x_k$ in $D_n$ converging to $a$ such that $x_k < a$. Fix $\epsilon < \frac{1}{4n}$. Then it is possible to find $\delta$ such that whenever $x\in A = (a-\delta, a) \cap E$,
$$|f(x)-f_-(a)| < \epsilon.$$
Choose an $x \in A$ and $x\in D_n$. Both $f_-(x)$ and $f_+(x)$ exist. Further it is possible to choose $\delta_1$ such that $x+\delta_1 < a$, $a-\delta < x- \delta_1$ and
$$|f(y)-f_-(x)| < \epsilon, \quad \forall y\in (x-\delta_1, x) \cap E$$
and
$$|f(y)-f_+(x)| < \epsilon, \quad \forall y\in (x, x+\delta_1) \cap E.$$
Let $y_1\in (x-\delta_1, x)\cap E$ and $y_2 \in (x, x+\delta_1)\cap E$. Then by our construction,
$$|f_-(x) - f_+(x)| = |f_-(x) - f(y_1) -f_-(a)+f(y_1) +f_-(a) - f(y_2) - f_+(x) + f(y_2)| \leq 4\epsilon < \frac {1}{n}.$$
So what we proved is that, if some element of $D_n$ is not isolated, then it is possible to have an element $x$ in $D_n$ such that $|f_-(x)  - f_+(x)| < \frac{1}{n}$.
A: 
How to show that every point of $D_n$ is an isolated point.

Suppose, to the contrary, that $a\in D_n$ has the property that $(a, a+\varepsilon)$ intersects $D_n$ for every $\varepsilon>0$.
This assumption means that $|f(b)-f_+(a)|>1/3n$ for at least one $b\in(a,a+\varepsilon)$ -- namely a point close enough to one of the sides of the jump of size $\ge 1/n$ we're assuming exists in $(a,a+\varepsilon)$.
Can you go the rest of the way now?
A: Suppose $x\in \overline {D_n\setminus \{x\} }$. Then WLOG let $(y_j)_{j\in \Bbb N}$ be a strictly increasing sequence of members of $D_n$ converging to $x$. For $j\ge 2$ there exists $u_j\in (y_{j-1},y_j)$ and $v_j\in (y_j,y_{j+1})$ such that $|f(u_j)-f(v_j)|>1/2n$ because $y_j\in D_n.$
Then $x\not\in D_n$
because $f_-(x)$ does not exist
because otherwise $f_-(x)=\lim_{j\to\infty}f(v_j)\ge 1/2n+\lim_{j\to\infty}f(u_j)=1/2n+f_-(x).$
Remark: If $x\in \overline {D_n\setminus \{x\} }$ there is  a strictly increasing or strictly decreasing sequence in $D_n$ converging to $x$. Here, "WLOG" ("without loss of generality") is mathspeak for "the other case is handled in the same manner and I assume this is obvious so do it yourself."
