I am searching an example of a continuous function $f: [0,1) \to [0,1)$ but f has no fixed point, that is, there is no point $x_0 \in [0,1)$ such that $f(x_0)\not= x_0 \forall x_0$.
Hint: consider polynomials of degree $1$ such that $f(1)=1$.
Always try to take motivation from picture.Draw straight line above line y=x which starts at (0,y) and ends at (1,1).y will be any value in (0,1) and you know equation of line passing through two points.
Such function does not exist.
Let $\phi(x)= f(x)-x$, $\phi(0)=f(0)\ge 0$ and $\phi(1)=f(1)-1<0$, therefore by ITV theorem there must be a $x_0\in [0;1)$ so that $\phi(x_0)=0$ which simply mean $f(x_0)=x_0$.
If you want a such function, you should avoid continuity, or the stability of $[0,1)$ by $f$.