Can we extend this proof for all $x$? I am not asking for any alternative proofs. I will be clear in what I am asking.

Theorem: Let $f:\mathbb{R}\to\mathbb{R}$ be a function over an interval
  $(a,b)$. If $f'(x)>0$, for every $x\in(a,b)$, then $f$ is increasing
  over $(a,b)$.

I will be posting a proof of this theorem for only a small interval.
Proof: Let $c\in(a,b)$. We know that $f'(c)>0$.Taking the right hand limit, we know that
$$\begin{align}&\lim_{x\to c^+}\frac{f(x)-f(c)}{x-c}=K>0\\
&\implies(\forall\epsilon>0)(\exists\delta>0)(0<x-c<\delta)(\left|\frac{f(x)-f(c)}{x-c}-K\right|<\epsilon)
\end{align}$$
Now, choose some $\epsilon<K$. And let $\delta$ be the respective number for this $\epsilon$. So, 
$$0<x-c<\delta\implies \frac{f(x)-f(c)}{x-c}\in(K-\epsilon,K+\epsilon)\implies\frac{f(x)-f(c)}{x-c}>0$$
And so, $f(x)>f(c)$. However, this proof only proves the inequality for $x<c+\delta$.
Initially, I thought that I could use these facts to prove for all $x\in(a,b)$, but I have an intuition that maybe this is impossible. So, I am interested to know if I can do it, and how.
NOTE:Please don't post other proofs of the theorem, I don't need them. Nor I need any advice on how to prove it. I only want info regarding these statements. 
 A: What you have shown is technically described by saying that $f$ is strictly increasing at $c$. The condition $f'(c) > 0$ is sufficient to show that $f$ is strictly increasing at $c$ (as you have shown in your post).
There is a further deep theorem based on completeness of real numbers which says that if a function is increasing at every point of an interval then it is increasing in the whole interval. The result is proved in this answer: https://math.stackexchange.com/a/2047743/72031
Another simpler but more technical approach is the use of Heine Borel Theorem on interval $[p, q] $ where $p, q\in (a, b) $ with $p<q$. For each point $x\in [p, q] $ we have a neighborhood $I_{x}$ of $x$ such that if $x>y, y\in I_{x} $ then $f(x) > f(y)$ and if $x<y, y\in I_{x}$ then $f(x) < f(y)$. By Heine Borel Theorem a finite number of such neighborhoods $I_{x}$ cover the interval $[p, q]$ and further these neighborhoods must overlap to cover $[p, q]$ and hence $f(p) < f(q) $.

The above two proofs show that the theorem

If $f'(x) > 0$ for all $x$ of some interval $I$ then $f$ is strictly increasing in $I$.

is a deep / difficult theorem. The relative simplicity of the proofs based on Rolle's theorem or Mean Value Theorem is an illusion because the proof of Rolle's theorem itself is very deep and ultimately rests on the completeness of real numbers.
See http://paramanands.blogspot.com/2012/07/monotone-functions-part-1.html and http://paramanands.blogspot.com/2012/07/monotone-functions-part-2.html for more details. 
A: Here's a proof sketch using an interesting technique that can be described as the intermediate value theorem applied to boolean predicates via the completeness of the real numbers. This technique can be used to prove a lot of theorems in analysis, including the intermediate value theorem, extreme value theorem and Dini's theorem.

Let $m \in (a,b)$.
Let $S = \{ x : x\in[m,b] \land \text{$f$ is increasing on $[m,x)$} \}$.
Note that $S$ is bounded above and non-empty since $m \in S$.
So let $c = \sup(S)$, which exists by completeness of $\mathbb{R}$.
Then it is simple to show that $f$ is increasing on $[m,c)$.
If $c < b$:
  Stick your proof here that shows that $f$ is increasing on $(c-d,c+d)$ for some $d > 0$.
  Since $[m,c)$ and $(c-d,c+d)$ overlap, it is easy to show that $f$ is increasing on $[m,c+d)$.
  But this contradictions the supremacy of $c$.
Therefore $c \ge b$ and hence $f$ is increasing on $[m,b)$.
By symmetry $f$ is increasing on $(a,m]$.
Since $(a,m]$ and $[m,b)$ overlap, $f$ is increasing on $(a,b)$.
