Let be $F:[a,b]\to\mathbb{R}$, where $F(x):=\int\limits_a^xf(t)dt$. We assume $f$ to be continuous at point $x_0$ and Riemann-integrable over $[a,b]$. Then we know that $F'(x_0)=f(x_0)$.
My (edited) approach (we assume $x<x_0$):
Let's define the function $M:[a,x_0]\to\mathbb{R}$, where $M(x):=\sup\{\left|f(t)-f(x_0)\right|\mid t\in[x,x_0]\}$.
If we choose an arbitrary $\frac{\epsilon}{2}>0$ then by continuity of $f$ at $x_0$ there exists a $\delta>0$ such that for all $t\in [x,x_0]$ with $|t-x_0|<\delta$ we have $|f(t)-f(x_0)|<\frac{\epsilon}{2}$. By definition of the supremum there exists a $t'\in[x,x_0]$ such that $M(x)-\frac{\epsilon}{2}<|f(t')-f(x_0)|$ and therefore $M(x)<\frac{\epsilon}{2}+|f(t')-f(x_0)|<\epsilon$. So if we only admit $x\in [a,x_0]$ such that $|x-x_0|<\delta$ we see that $M(x)$ is continuous at $x_0$.
Keeping $|x-x_0|<\delta$ in mind, allows us to conclude:
$$
\left|\frac{\int\limits_x^{x_0}f(t)dt}{x-x_0}-f(x_0)\right|=\left|\frac{\int\limits_x^{x_0}f(t)-f(x_0)dt}{x-x_0}\right|\leq \left|\frac{M(x)(x_0-x)}{x-x_0}\right| = M(x) <\epsilon.
$$
The case "$x>x_0$" is basically is the same. Hence, $F'(x_0)=f(x_0)$.
Is this proof o.k.?
Best Answer
You are fairly close, but as I note in the comments, there are a few problems with what you write. There is an additional issue at the end. I had already written up a very similar approach, which is how I usually do it, but let me address how to fix your argument:
Let us assume $f(x)$ is integrable on $[a,b]$ (with $x_0\in (a,b)$) so that $F(x)$ makes sense.
To show continuity of $M(x)$ from the left at $x_0$ (equivalently, show its limit as $x\to x_0^-$ is $0$), proceeding as you do until you pick $t'$ (which was a problem prior to the edits as I noted in comments), let us do this instead: we know there exists $t'\in [x,x_0]$ such that $$M(x) - \frac{\epsilon}{2} \lt |f(t')-f(x_0)| \leq M(x).$$ Hence $$0\leq M(x) \lt |f(t')-f(x_0)|+\frac{\epsilon}{2} \lt \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon.$$ This holds for all $\epsilon\gt 0$, so we conclude the limit is indeed $0$.
Finally, there is a small correction I think you need to make to the final chain of inequalities, namely $r\leq s$ does not imply that $|\int_x^{x_0} r\,dt| \leq |\int_x^{x_0}s\,dt|$. (For example, integrade $-3$ on $[0,1]$, compraed to the integral of $-1$ on $[0,1]$).
Instead, you should remember that in general $$\left|\int_a^b f(t)\,dt\right| \leq \int_a^b |f(t)|\,dt,$$ assuming both integrals exist, so that you can say $$\begin{align*} \left|\frac{\int_x^{x_0}f(t)\,dt}{x-x_0} - f(x_0)\right| &= \left|\frac{\int_x^{x_0}(f(t)-f(x_0))\,dt}{x-x_0}\right|\\ &\leq \frac{\int_x^{x_0}|f(t)-f(x_0)|\,dt}{|x-x_0|}\\ &\leq \frac{\int_x^{x_0}M(x)\,dt}{|x-x_0|}\\ &= \frac{M(x)(x_0-x)}{x_0-x} = M(x), \end{align*}$$ and then taking the limit as $x\to x_0^-$ you get the desired $$\lim_{x\to x_0^-}\frac{\int_x^{x_0} f(t)\,dt}{x-x_0} = f(x_0).$$ And then a similar correction and fix is needed for the continuity from the right and the derivative from the right.
As I had already written up the stuff below, I keep it here. This is how I usually do this.
Rather than work with the supremum of the absolute value of the difference on the interval, I'm just going to work with the supremum and infimum of the values on that interval. It should be straightforward (though wordy) to convert this to your function.
I will restrict to $x\leq x_0$, as you do. Let me define functions that are close to your $M(x)$: for $x\lt x_0$, $$\begin{align*} m(x) &= \inf\{ f(t)\mid x\leq t\leq x_0\}\\ M(x) &= \sup\{ f(t)\mid x\leq t\leq x_0\}. \end{align*}$$ Because $f(x)$ is integrable on $[x,x_0]$, we know it is bounded, so $m(x)$ and $M(x)$ both make sense.
I claim that $$\lim_{x\to x_0^-}(M(x)-m(x)) = 0.$$ Indeed: let $\epsilon\gt 0$. Since $f$ is continuous at $x_0$, we know that there exists $\delta\gt 0$ such that if $t\in [a,x_0]$ and $|t-x_0|\lt\delta$, then $|f(t)-f(x_0)|\lt \frac{\epsilon}{4}$. Shrinking $\delta$ if necessary, we may assume that $a\leq x_0-\delta$, so that if $0\leq x_0-t\lt\delta$, then $t$ is in $[a,x_0]$.
Let $x$ be such that $0\leq x_0-x\lt\delta$. Since $M(x)$ is a supremum, there exists $t_M\in [x,x_0]$ such that $M(x)-\frac{\epsilon}{4}\lt f(t)\leq M(x)$; and there exists $t_m\in [x,x_0]$ such that $m(x)\leq f(t_m)\lt m(x)+\frac{\epsilon}{4}$. Then $$\begin{align*} |M(x)-m(x)| &= |M(x) - f(t_M) + f(t_M) -f(x_0)+f(x_0)- f(t_m) + f(t_m)-m(x)|\\ &\leq |M(x)-f(t_M)| + |f(t_M)-f(x_0)| \\ &\qquad\mathop{+} |f(x_0)-f(t_m)| + |f(t_m)-m(x)|\\ &\lt \frac{\epsilon}{4} + \frac{\epsilon}{4} + \frac{\epsilon}{4} + \frac{\epsilon}{4}\\ &= \epsilon, \end{align*}$$ where the first $\frac{\epsilon}{4}$ follows from $M(x)-\frac{\epsilon}{4}\lt f(t_M)$; the second from the continuity of $f$ at $x_0$ and the fact that $|t_M-x_0|\lt\delta$; the third from continuity of $f$ at $x_0$ and the fact that $|t_m-x_0|\lt\delta$; and the fourth from the fact that $f(t_m)\lt m(x)+\frac{\epsilon}{4}$.
Thus, the limit of $M(x)-m(x)$ is $0$ as $x\to x_0^-$.
(This argument can be simplified if you assume $f$ is continuous on the entire interval and you take the Extreme Value Theorem for granted, since then we can actually achieve $M(x)$ and $m(x)$, so we don't have to approximate them.)
Likewise, since $m(x)\leq f(x_0)\leq M(x)$, we have $$\begin{align*} 0&\leq \lim_{x\to x_0^-}(f(x_0)-m(x))\\ &\leq \lim_{x\to x_0^-}(M(x)-m(x))\\ &= 0, \end{align*}$$ so $\lim_{x\to x_0^-}m(x) = f(x_0)$; similarly, $\lim_{x\to x_0^-}M(x)=f(x_0)$.
So now, $$\lim_{x\to x_0^-}\frac{F(x)-F(x_0)}{x-x_0} = \lim_{x\to x_0^-}\frac{1}{x-x_0}\int_{x_0}^x f(t)\,dt.$$ Since $m(x)\leq f(t)\leq M(x)$ for all $t\in [x,x_0]$, $$m(x) \leq \frac{1}{x-x_0} \int_{x_0}^xf(t)\,dt \leq M(x).$$ Therefore, $$0 \leq \left(\frac{1}{x-x_0}\int_{x_0}^x f(t)\,dt\right) - m(x) \leq M(x)-m(x).$$ Letting $x\to x_0^{-}$, the Squeeze Theorem tells us that $$\lim_{x\to x_0^-}\left(\frac{1}{x-x_0}\int_{x_0}^x f(t)\,dt\right) - m(x) = 0.$$ Since $\lim_{x\to x_0^-}m(x) = f(x_0)$, then $$\lim_{x\to x_0^-}\frac{1}{x-x_0}\int_{x_0}^x f(t)\,dt$$ exists, and equals $\lim_{x\to x_0^-}m(x) = f(x_0)$, as required.
A similar argument works if $x_0\leq x\leq b$.