If $f$ is differntiable at $x_0$, then $f$ is continuous at $x_0$. How does this work? There's a similar post here. but I haven't got the answer: 
If $f$ is differentiable at $x = x_0$ then $f$ is continuous at $x = x_0$.
I just started learning calculus for physics, so I don't understand well yet. I was watching an MIT OpencourseWare Lecture (link below), and I just couldn't understand the last section of the video where the proof of a theorem was being explained. I did search online for answers but didn't understand their explanations, and it feels like I'm missing something really obvious.
The theorem states: If $f$ is differentiable at $x_0$, then $f$ is continuous. The proof goes like this:
1)
$$
\lim_{x\to x_0} f(x)-f(x_0) = 0
$$
2)
$$
\lim_{x\to x_0} \dfrac{f(x)-f(x_0)}{x-x_0}\cdot(x-x_0)=f'(x_0)\cdot0=0
$$
What did I understand is that the first equation is just the definition of continuity: $\lim\limits_{x\to x_0} f(x) = f(x_0)$ rearranged. But I do not understand the second part at all. 

Can someone please explain what is happening in the second equation? Also: 
1) Where does the $f'(x_0)$ suddenly come from? 
2) Why is it being multiplied with $\dfrac{x-x_0}{x-x_0}$?
It does look vaguely familar to the difference quotient: $\lim\limits_{\Delta x\to 0} f'(x_0) = \dfrac{f(x_0+\Delta x) - f(x_0)}{\Delta x}$
I am suspecting that it has something to do with $\Delta x$ being implied somewhere.
Here are the links:
Video (starts at 46:10)
PDF version
 A: Your definition is almost correct: it should be 
$$f^{\prime}(x_0) = \lim_{\Delta x \to 0}\dfrac{f(x_0+\Delta x) - f(x_0)}{\Delta x}\text{.}$$
To see how this definition can be written differently, perform a change of variables, where $x = x_0 + \Delta x$. Then the limit is taken as $\Delta x = x - x_0 \to 0$ or equivalently $x \to x_0$, hence
$$f^{\prime}(x_0) = \lim_{x - x_0 \to 0}\dfrac{f(x) - f(x_0)}{x - x_0} = \lim_{x \to x_0} \dfrac{f(x) - f(x_0)}{x - x_0}\text{.}$$
Now as $x \to x_0$, $x - x_0 \to 0$. Hence
$$\lim_{x \to x_0} \left[\dfrac{f(x) - f(x_0)}{x - x_0}\cdot (x-x_0)\right] = \lim_{x \to x_0} \left[\dfrac{f(x) - f(x_0)}{x - x_0}\right]\cdot \lim_{x \to x_0}(x-x_0) = f^{\prime}(x_0) \cdot 0\text{.}$$
A: The proof is very badly laid out. (Are you sure you copied it right?) It should go like this:
$$\lim_{x\to x_0} (f(x)-f(x_0)) = \lim_{x\to x_0} \dfrac{f(x)-f(x_0)}{x-x_0}\cdot(x-x_0)=f'(x_0)\cdot0=0$$
where we have used the fact that the limit of a product is the product of the limits, if both exist. Also we have substituted $x-x_0$ for your $\Delta x$ in the definition of the derivative. From this we have
$$\lim_{x\to x_0} (f(x)-f(x_0)) = 0$$
which is the definition of continuity at $x_0$.
