Question: I don't understand why this is true:
If $f:[0,1]\rightarrow[0,2]$ is a continuous function then exists $x \in [0,1]$ such that $f(x)=2x$

I don't understand why such a point exist. Why is there not for example a fixpoint $f(x)=x$?

Answer: 
There does not *have* to be a fixed point $x$ such that $f(x) = x$. Here's a counterexample: the constant function $f(x) = 2$. Obviously, there is no $x \in [0, 1]$ for which $x = 2$.

Given a function $f(x)$ as described in the problem, consider the function $g(x) = f(x)-2x$. We have $g(0) \in [0, 2]$, and $g(1) \in [-2, 0]$. Since $g(x)$, like $f(x)$, is continuous, there must—by way of the intermediate value theorem—exist a value $x \in [0, 1]$ such that $g(x) = 0$. For that value of $x$, we must have $f(x) = 2x$.

More broadly, the reason why there must be an $x \in [0, 1]$ such that $f(x) = 2x$ is that the line $y = 2x$ is a diagonal of the rectangle $[0, 1] \times [0, 2]$. You start off on the left side of the rectangle, and end up on the right side, and in so doing, must cross the diagonal. The other diagonal also works this way: One might just as effectively say that there must be an $x \in [0, 1]$ such that $f(x) = 2-2x$. Hopefully, you can now see why that must be so.
Why is there not a fixed point $f(x)=x$? Nobody said there wasn't.

To show there does exist an $x$ with $f(x)=2x$, let $g(x)=f(x)-2x$. Then $g(0)=f(0)\ge 0$, while $g(1)=f(1)-2\le0$. So the intermediate value theorem shows there exists $x$ with $g(x)=0$.

(Or, if you have a theorem saying any map from $[0,1]$ to itself has a fixed point, consider $f/2$.)