Proof of first Fundamental theorem of calculus Can you please, check if my proof is correct?

Suppose that $f:[a,b]\to \Bbb{R}$ is continuous and $F(x)=\int^{x}_{a}f(t)dt$, then  $F\in C^{1}[a,b]$ and 
  $$\dfrac{d}{dx}\int^{x}_{a}f(t)dt:=F'(x)=f(x)$$

MY PROOF: Credits to Aweygan for the correction
Let $x_0\in[a,b]$ and $\epsilon>0$ be given. Since $f$ is continuous at $x_0$ then, there exists $\delta>0$ such that $|t-x_0|<\delta$ implies $$|f(t)-f(x_0)|<\epsilon.$$
Thus,  $$f(x_0)=\dfrac{1}{x-x_0}\int^{x}_{x_0}f(x_0)dt,\;\;\text{where}\;\;x\neq x_0.$$
For any $x\in (a,b),$ with $0<|x-x_0|<\delta,$ such that $x_1=\min\{x,x_0\}$ and $x_2=\max\{x,x_0\}$. So, we have
\begin{align}\left| \dfrac{F(x)-F(x_0)}{x-x_0}-f(x_0) \right|&=  \left| \dfrac{1}{x-x_0}\int^{x}_{x_0}(f(t)-f(x_0))dt \right|  \\&\leq  \dfrac{1}{|x-x_0|}\int^{x}_{x_0} \left|f(t)-f(x_0) \right|dt\\&\leq  \dfrac{1}{|x-x_0|}\int^{x_2}_{x_1} \left|f(t)-f(x_0) \right|dt\\&< \dfrac{1}{|x-x_0|}\epsilon|x_1-x_2| \\&\leq \dfrac{1}{|x-x_0|}\epsilon|x-x_0| =\epsilon  \end{align}
Hence, 
$$F\in C^{1}[a,b]\;\;\text{and}\;\;\dfrac{d}{dx}\int^{x}_{a}f(t)dt:=F'(x)=f(x)$$

It's essentially correct, but you should either split up the last part of the proof into the cases where $x<x_0$ and $x_0<x$, or write $x_1=\min\{x,x_0\}$, $x_2=\max\{x,x_0\}$ and do the following:
\begin{align}
\left| \dfrac{F(x)-F(x_0)}{x-x_0}-f(x_0) \right|&=  \left| \dfrac{1}{x-x_0}\int^{x}_{x_0}(f(t)-f(x_0))dt \right|  \\
&\leq  \dfrac{1}{|x-x_0|}\int^{x_2}_{x_1} \left|f(t)-f(x_0) \right|dt
\\&< \dfrac{1}{|x-x_0|}\epsilon|x-x_0| \\
&=\epsilon.  
\end{align}
