Given that $f:Df\rightarrow R, Df \subset R, a \in \overline R , a\in D'f, b\in \overline R, c\in \overline R$ and $\lim_{x\to a}f(x)=b$ and $\lim_{x\to a}f(x)=c$ then $b=c$.
There are two things I don't get about the proof, first one is: If $a\in D'f$ then $\exists (x_n)$ such that $(\forall n\in N \ \ x_n\in Df\setminus\{a\})$ and $\lim_{n \to \infty}x_n=a$.
Why does $a$ being a limit point implies that there must be a sequence that converges to it? And that the sequence is within the domain? I could imagine that the definition of limit point implies that some sequence must converge to it, but I still don't know why that is.
Second problem: $\lim_{x \to a}f(x)=b \Rightarrow \lim_{n \to \infty}f(x_n)=b$ and $\lim_{x \to a}f(x)=c \Rightarrow \lim_{n \to \infty}f(x_n)=c$
Why is that, is there any proof or axiom for it?
Judging from comments you are dealing with metric spaces. So I will assume that, although the proof works for the topological setting as well (with some minor tweaks).
So say $(X,d)$ is our metric space, $D$ is the set in question and $a$ is its limit point, meaning for any $\epsilon>0$ there is $x\in D, x\neq a$ such that $d(x,a)<\epsilon$.
Now consider a sequence of real numbers $\epsilon_n=\frac{1}{n}$ and apply the definition to each $\epsilon_n$. So for each $\epsilon_n$ we have $x_n$ such that $d(a,x_n)<\epsilon_n=\frac{1}{n}$. By the Axiom of (Countable) Choice we can combine these elements into a proper sequence $(x_n)$. And I hope it is clear that the sequence has desired properties.