Monotonic function on $\mathbb{R}$ It is a well known fact, that if $f:X \to \mathbb{R}$ (where $X \subset \mathbb{R}$) is monotonic, and $a \in X^+, b\in X^-$, where 
$$X^+ = \left\{ {x\in X:\forall \varepsilon  > 0,\;\;\left( {x - \varepsilon ,x} \right) \cap X \ne \varnothing} \right\}$$
$$X^- = \left\{ {x\in X:\forall \varepsilon  > 0,\;\;\left( {x,x + \varepsilon } \right) \cap X \ne \varnothing} \right\},$$
then
$$\lim_{x \to a^+} f\left( x \right),\qquad\lim_{x \to b^-} f\left( x \right)$$
exist. My question is, given such a function, and a point $a \in X^+$, does there exist an  $\varepsilon > 0$ such that $$
f\left| {_{\left( {x - \varepsilon ,x} \right) \cap X} } \right.
$$
is continuous (doesn't have jumps). I want to know if this fact is true, to prove that the set of discontinuities of a monotonic function is countable (because in this case, to each such point, I associate to it, an open set, and therefore a rational number).
The only possible counterexample is a function that has a dense set of jumps. Please don't give me proof about the other proof, I only know this, because I know other proof without using this fact, but I think that this is also true.
 A: I think you have your definitions of $X^+$ and $X^-$ the wrong way around: under your definition, if $X=[0,1]$, then $1\in X^+$, but $\lim\limits_{x\to 1^+} f(x)$ does not make sense, so it cannot exist.
Your question has a negative answer.
Let $\{q_n\}$ be an enumeration of the rationals on $[0,1]$; that is, $q_n$ is a rational on $[0,1]$ for each $n$, and for every rational $q$ in $[0,1]$ there exists one and only one $n$ such that $q=q_n$.
Define $f\colon [0,1]\to \mathbb{R}$ by
$$f(x)= \sum_{q_n\gt x}\frac{1}{2^n}.$$
Then $f(x)$ is monotonic: if $x\leq y$, then $\{n\in\mathbb{N}\mid q_n\gt y\}\subseteq \{n\in\mathbb{N}\mid q_n\gt x\}$, so $f(x)\geq f(y)$. 
Moreover, $f$ is discontinuous from the left at every rational number: if $q\in (0,1]\cap\mathbb{Q}$, there exists $n$ such that $q=q_n$. Let $\epsilon=\frac{1}{2^{n+1}}$. Then for every $x\in(0,q)$, $f(x)-f(q) \geq \frac{1}{2^n}\gt \epsilon$, so $f$ is not continuous from the left at $q$. 
In particular, for every $x$, $1\geq x\gt 0$ and every $\epsilon\gt 0$, $f$ is not continuous on $(\max(x-\epsilon,0),x)$.
The function is continuous from the right at every rational, but it is likewise true that for all $x$, $ 0\leq x\lt 1$, and for every $\epsilon\gt 0$, $f$ is not continuous on $(x,\min(x+\epsilon,1))$. 
Note that in this case, $X^+$ and $X^-$ are $[0,1)$ and $(0,1]$ (in some order, depending on the correct definitions of $X^+$ and $X^-$).
A: Proposition 1. A monotone function $f: \mathbb{R} \to \mathbb{R}$ can only have jump discontinuities.
Corollary 2. A monotone function $f: \mathbb{R} \to \mathbb{R}$ has at most countably many discontinuities.
Proposition 3. For any countable $A \subset \mathbb{R}$ there exists a bounded increasing function $f: \mathbb{R} \to \mathbb{R}$ such that $f$ is continuous precisely on $\mathbb{R}\setminus A$.
Only the last statement is moderately difficult to prove.
