A monotonous function $f:I\rightarrow \mathbb{R}$ on an Intervall $I$ has at most countable (bijective to $\mathbb{N}$) many uncontinious points I have first tried to prove the Statement by contraposition but I don't know how I can continue on a certain step. I have also read the Solutions which I don't understand.
$(x_n)$ is a sequence of not continous/uncontinious Points. The sequence is bounded and the point a subsequence converges must also be in the interval.
Let $L$ be this Point. Don't know what I can do from here on.
The solution said:
Suppose $f$ is increasing. There are $f(x+)$ and $f(x-)$ for every $x\in I$. Due to monotonicity $f(x+)\geq f(x-)$ with equality if $f$ is continious in $x$. $U$ is the set of all the Points/ x's which are uncontinious. We choose for every $x\in U$ a rational number $r(x)$ between $f(x-)$ and $f(x+)$. The map $r:U\rightarrow \mathbb{Q}$ is injective because $f$ is monotonous. $\mathbb{Q}$ is countable therefore $U$ is at most countable
I don't know what he meant with $f(+)$ and $f(-)$, does he mean the right and left Limit for $x$? Or does he just mean a pair of Points where one is right and the other one is left from $x$?. The Limit does not make much sense but on the other Hand if it is a pair of Points then why $x$ is necessarily continious if $f(x+)=f(x-)$? I think it is a pair of Points if I have missunderstood that part please correct me.
We take a uncontinious Point $x$ and we choose a rational Point $r(x)$ for which we have $f(x-)\leq f(r(x))\leq f(x+)$, why do we do that? I still don't know what $f(x+)$ and $f(x-)$ is
 A: The notation $f(x+)$ is standard for the lateral (right) limit $\lim_{h\to 0^+}f(x+h)=\lim_{z\to x^+}f(z)$ (whenever it exists); Similarly, $f(x-)$ is the left limit.
Let $x$ be temporarily fixed. Since $f$ is increasing, then $f(x+h)\geq f(x)$ whenever $h>0$. This proves that $\left\{f(x+h):h>0\right\}$ is bounded below (by $f(x)$) and thus has an infimum $L(x)$, which satisfies $f(x)\leq L(x)$. You can show that $L(x)=f(x+)$, i.e., that for all $\epsilon>0$ there exists $\delta>0$ such that $0<h<\delta$ implies
$$|f(x+h)-L(x)|<\epsilon$$
(this is simply the definition of the lateral right limit.)
Similarly, $f(x-)$ always exists: it is $\sup\left\{f(x+h):h<0\right\}$, and it satisfies $f(x-)\leq f(x)$.
Now you can recall a basic theorem from calculus:

The function $f$ is continuous at $x$ if and only if $f(x-)=f(x)=f(x+)$;

But since in this case we already know that $f(x-)\leq f(x)\leq f(x+)$, we can rewrite this as:

The function $f$ is continuous at $x$ if and only if $f(x+)\leq f(x-)$.

The proof goes by assuming that $x$ is a discontinuity point. By the (contrapositive of the) statement in the box above, this means that $f(x-)<f(x+). Then we take any rational $r(x)$ satisfying $f(-)

Now you need to prove that the function $x\mapsto r(x)$ is injective. In fact, it is increasing in its domain (it is only defined for discontinuity points): Suppose $x<y$. Since $f(x+)$, the lateral limit, only cares about points very close to $x$, then in fact we can modify the $f(x+)=\inf\left\{f(z):z>x\right\}$ that we obtained above to $f(x+)=\inf\left\{f(z):x<z<y\right\}$. Similarly, $f(y-)=\sup\left\{f(z):x<z<y\right\}$, so $f(x+)\leq f(y-)$. If both $x$ and $y$ are discontinuity points, then
$$r(x)< f(x+)\leq f(y-)<r(y).$$
so the function $r$ is strictly increasing and in particular one-to-one.
A: You're correct in your interpretation: $$f(x^+)=\lim_{t\searrow x}f(t)$$ and $$f(x^-)=\lim_{t\nearrow x}f(t)$$ are respectively the right-hand and left-hand limits of $f$ at $x.$
Since $f$ is monotonic and increasing, then for $x$ in the interior of $I,$ we have  $$f(x^-)\le f(x)\le f(x^+),$$ so $f$ is continuous at $x$ if and only if $f(x^-)=f(x^+).$
Thus, if $f$ has an internal discontinuity at $x,$ then we'll have $f(x^-)<f(x^+).$ Since the rational numbers are dense in the reals, then there is a rational number $r(x)$ such that $f(x^-)<r(x)<f(x^+).$ Here, $r(x)$ isn't a domain value, nor necessarily even a range value. Rather, it is just a rational number inside the "gap" of the jump discontinuity.
Since the map $x\mapsto r(x)$ is injective, then $f$ has at most countably-infinitely-many internal discontinuities, and at most $2$ non-internal discontinuities (at the endpoints of $I$, if they exist and $f$ is defined there). Thus, there are at most countably-infinitely-many discontinuities.
The fact that $I$ is an interval turns out to be immaterial to the result. See here for more.
