Question: > 
> Claim: if $f$ is differentiable at $x = x\_0$ then $f$ is continuous at $x = x\_0$.
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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.

Answer: 
$$\lim\limits\_{x\to x\_0}f(x)=\lim\limits\_{x\to x\_0}\left(f(x\_0)+(x-x\_0)\cdot\dfrac {f(x)-f(x\_0)}{x-x\_0}\right)=$$
$$\lim\limits\_{x\to x\_0}f(x\_0)+\lim\limits\_{x\to x\_0}(x-x\_0)\cdot\lim\limits\_{x\to x\_0}\dfrac {f(x)-f(x\_0)}{x-x\_0}=$$
$$f(x\_0)+0\cdot f'(x\_0)=f(x\_0)$$
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 $x\_0$.