The Problem : Let $J \subseteq \mathbb{R}$ be an interval and $f : J \to \mathbb{R}$ be increasing. Define the jump at the point $c$ as follows:
$j_f(c)=\lim_{x \to c+}f(x)-\lim_{x \to c-}f(x)$ if $c \in J$ is not an end-point of $J.$
$j_f(c)=\lim_{x \to c+}f(x)-f(c)$ if $c \in J$ is the left end-point of $J.$
$j_f(c)=f(c)-\lim_{x \to c-}f(x)$ if $c \in J$ is the right end-point of $J.$
To show that $f$ is continuous at $c \in J$ iff $j_f(c)=0.$
I'm attaching my solution (A few pieces of argument is marked as informal as they are similar to the previous line of argument and hence not elaborated). Please notify if there's any gap/flaws in the arguments, also whether it could be made shorter by any other technique. Any comment/suggestion regarding this proof, or maybe in general context (style of proof-writing etc), would be greatly appreciated. Thank you.
My Solution :
$(\Rightarrow)$ part : Suppose $c$ is not an end-point of $J$. Given $f$ is continuous at $c.$ Fix $\epsilon > 0.$ Then $\exists \delta > 0$ such that $x \in (c-\delta,c+\delta) \cap J \Rightarrow |f(x)-f(c)| < \epsilon.$ Hence $$x \in (c-\delta,c) \cap J \Rightarrow |f(x)-f(c)| < \epsilon \tag 1$$ $$x \in (c,c+\delta) \cap J \Rightarrow |f(x)-f(c)| < \epsilon \tag 2$$ Since $\epsilon > 0$ is arbitrary, we conclude from $(1)$ that $\lim_{x \to c-}f(x)=f(c)$ and from $(2)$ that $\lim_{x \to c+}f(x)=f(c).$ Hence $j_f(c)=\lim_{x \to c+}f(x)-\lim_{x \to c-}f(x)=f(c)-f(c)=0.$
Now suppose $c$ is the left end-point of $J$. Given $f$ is continuous at $c.$ Then $\exists \delta > 0$ such that $x \in (c-\delta,c+\delta) \cap J = [c,c+\delta) \cap J \Rightarrow |f(x)-f(c)| < \epsilon.$ So, $x \in (c,c+\delta) \cap J \Rightarrow |f(x)-f(c)| < \epsilon.$ Thus $\lim_{x \to c+}f(x)=f(c),$ and hence $j_f(c)=0.$
Informal : Similarly, if $c$ is the right-end point of $J,$ we obtain $j_f(c)=0.$
Note : We did not need the fact that $f$ is an increasing function to establish this side of the equivalence.
$(\Leftarrow)$ part : Suppose $c$ is not an end-point of $J.$ Now $j_f(c)=0 \Leftrightarrow \lim_{x \to c-}f(x)=\lim_{x \to c+}f(x)=l$ $($let$).$ Thus $\exists \delta_1>0$ and $\delta_2>0$ such that
$$x \in (c-\delta_1, c) \cap J \setminus \{c\} \Rightarrow |f(x)-l|<\epsilon$$ $$x \in (c, c+\delta_2) \cap J \setminus \{c\} \Rightarrow |f(x)-l|<\epsilon$$ Let $\delta=min\{\delta_1, \delta_2\}>0$. Then $$x \in (c-\delta, c+\delta) \cap J \setminus \{c\} \Rightarrow |f(x)-l|<\epsilon \tag 3$$
Claim : $l=f(c).$
We prove the claim by contradiction. First we assume $l<f(c).$ Choose $\epsilon = \frac{f(c)-l}{2}>0.$ Then $\exists \delta_3>0$ such that $x \in (c-\delta_3,c+\delta_3) \cap J \setminus \{c\} \Rightarrow |f(x)-l|<\epsilon,$ in particular $f(c+\frac{\delta_3}{2})<l+\epsilon=\frac{l+f(c)}{2}<f(c),$ which is a contradiction since $f$ is increasing.
Informal : If otherwise, let $l>f(c),$ we similarly choose $\epsilon=\frac{l-f(c)}{2}>0$ and a corresponding $\delta_4>0.$ Check that $f(c-\frac{\delta_4}{2})>f(c),$ contradicting the fact that $f$ is increasing.
Hence we conclude (by trichotomy law of real numbers) that $l=f(c).$ Now, putting $l=f(c)$ in $(3)$ and noting that $x=c \Rightarrow |f(x)-f(c)|=0<\epsilon,$ we obtain
$$x \in (c-\delta, c+\delta) \cap J \Rightarrow |f(x)-f(c)|<\epsilon \tag 4$$
Since $\epsilon>0$ is arbitrary, we conclude that $f$ is continuous at $c.$
Now suppose $c$ is the left end-point of $J.$ Then $j_f(c)=0 \Leftrightarrow \lim_{x \to c+}f(x)=f(c),$ i.e. given $\epsilon>0, \exists \delta > 0,$ such that $x \in (c, c+\delta) \cap J \Rightarrow |f(x)-f(c)|<\epsilon.$ At $x=c,$ obviously $|f(x)-f(c)|=0<\epsilon.$ Thus $x \in (c-\delta, c+\delta) \cap J \Rightarrow |f(x)-f(c)|<\epsilon,$ and hence $f$ is continuous at $c.$
Informal : Similarly if $c$ is the right end-point of $J,$ we have $j_f(c)=0 \Rightarrow$ $f$ is continuous at $c.$
This completes the proof. $\square$
Your proof seems correct! Also You could simply say since $f$ is increasing then $$L(c):= \lim_{x \to c-}f(x) \leq f(c) \leq \lim_{x \to c+}f(x)=R(c) $$ (for end points one of these two limit is same as $f(c)$) Now, under the assumption that $f$ is increasing $$j_f(c)=0 \quad \text{iff} \quad L(c)=R(c)=f(c) \quad \text{iff} \quad\text{ f is continuous at $c$}$$