I've tried to prove the theorem in advance at the level that satisfies me. The notation used might not be correct, but I hope all major steps are correct.
DEFINITION (from Apostol's Calculus I):
Let $f$ be a function that is integrable on $[a,x]$, for each $x$ in $[a,b]$. Let $c$ be such that $a \le c \le b$ and define a new function $A$ as follows: $$A(x) = \int_c^xf(t)dt, a \le x \le b$$ Then the derivative $A'(x)$ exists at each point $x$ in the open interval $(a,b)$ where $f$ is continuous, and for such $x$ we have $$A'(x) = f(x).$$
PROOF:
I will show that the theorem is a consequence of the property of continuity of the function at some point.
First we make some derivations. For simplicity lets assume that $h$ is positive. The proof for negative $h$ is no harder.
$$A'(x) = \lim_{h\to0} \frac{A(x+h) - A(x)}{h} = \lim_{h\to0}\frac{\int_c^{x+h}f(t)dt - \int_c^{x}f(t)dt}{h} = \lim_{h\to0}\frac{\int_x^{x+h}f(t)dt}{h}$$
Since $f$ is continuous at $x$, for any number $\epsilon$ there is some $\delta$, such that when $t \in [x-\delta, x+\delta]$,
$$f(x) - \epsilon \le f(t) \le f(x) + \epsilon$$
Then for any $h < \delta$,
$$\int_x^{x+h} [f(x) - \epsilon]dt \le \int_x^{x+h} f(t)dt \le \int_x^{x+h} [f(x) + \epsilon]dt$$
$$[f(x) - \epsilon]h \le \int_x^{x+h} f(t)dt \le [f(x) + \epsilon]h$$
$$f(x) - \epsilon \le \frac {\int_x^{x+h} f(t)dt}{h} \le f(x) + \epsilon$$
I can make $\epsilon$ as small as possible, hence
$$\lim_{h\to0}\frac{\int_x^{x+h}f(t)dt}{h} = A'(x) = f(x)$$
