If $f$ is differentiable at $x = x_0$ then $f$ is continuous at $x = x_0$. 
Claim: if $f$ is differentiable at $x = x_0$ then $f$ is continuous at $x = x_0$.

Please, see if I made some mistake in the proof below. I mention some theorems in the proof: 
The condition to $f(x)$ be continuous at $x=x_0$ is $\lim\limits_{x\to x_0} f(x)=f(x_0)$.
(1) If $f(x)$ is differentiable at $x-x_0$, then $f'(x)=\lim\limits_{x\to x_0} \dfrac{f(x)-f(x_0)}{x-x_0}$ exists and the function is defined at $x=x_0$. 
(2) Therefore, by the Limit Linearity Theorem, $\lim\limits_{x\to x_0} f(x)$ exists and we'll show it is equals $f(x_0)$. 
(3) We'll do this by the Precise Limit Definion: given $ \epsilon>0, \exists\delta|0<|x-x_0|<\delta$, then $0<|f(x)-f(x_0)|<\epsilon$. As this limit exists by (2), we can make $f(x)$ as close to  $f(x_0)$ as one wishes, therefore $\lim\limits_{x\to x_0} f(x)=f(x_0)$, what satisfies the condition for $f(x)$ be differentiable at  $x=x_0$. The end.
 A: You surely mess up things in the proof. You say we can make $f(x)$ as close to $f(x_0)$ as one whishes. But this is exactly what you are trying to prove.
The more useful definition of differentiability is, that a function is differentiable if
$f(x)=f(x_0) + a \cdot (x-x_0) + r(x)$ with 
$$\lim_{x\to x_0} \frac{r(x)}{x-x_0} = 0 $$
so $\lim_{x\to x_0} r(x)$ will surely be $0$. 
Hence 
\begin{align*}
|f(x)-f(x_0)|&= | f(x_0) + a \cdot (x-x_0) + r(x) -f(x_0)|\\
&\leq |a| \cdot |x-x_0|+ |r(x)|
\end{align*}
Now you have a sum of 2 terms which go to zero as $x$ goes to $x_0$.
A: $$\lim\limits_{x\to x_0}f(x)=\lim\limits_{x\to x_0}\left(f(x_0)+(x-x_0)\cdot\dfrac {f(x)-f(x_0)}{x-x_0}\right)=$$
$$\lim\limits_{x\to x_0}f(x_0)+\lim\limits_{x\to x_0}(x-x_0)\cdot\lim\limits_{x\to x_0}\dfrac {f(x)-f(x_0)}{x-x_0}=$$
$$f(x_0)+0\cdot f'(x_0)=f(x_0)$$
A: Here, a solution I've found at "Introduction to Analysis" byt Arthur Mattuck:
$$\lim_{x\to x_0} (f(x)-f(x_0)) = \lim_{x\to x_0} (\dfrac{f(x)-f(x_0)}{x-x_0})(x-x_0)\\ =  \lim_{x\to x_0} (\dfrac{f(x)-f(x_0)}{x-x_0}) \cdot \lim_{x\to x_0} (x-x_0)\\
=  f'(x)\cdot 0 \\
=0$$
