If $f$ is a homeomorphism then any periodic point have period less or equal 2 How can one prove the following statement?
Let $f:[0,1]\to [0,1]$ be a homeomorphism. If $x\in\operatorname{Per}(f)$ then the period of $x$ can't be greater than $2$, i.e, $f(x)=x$ or $f^2(x)=x$.
 A: $f$ must be strictly monotonic, otherwise it cannot be bijective and continuous. 
If $f$ is a homeomorphism, then so is $f^n$.
If $f$ is strictly increasing, then any periodic point must have period 1. If not, and $x$ is a fixed point of $f^n$ (but not $f^{n-1}$), then either $x< f(x) < f^2(x) < \cdots < f^n(x) = x$, or
$x> f(x) > f^2(x) > \cdots > f^n(x) = x$ which gives a contradiction.
So, we can assume that  $f$ is strictly decreasing.
Then $f^{2n}$ is strictly increasing and $f^{2n+1}$ is strictly decreasing.
The intermediate value theorem shows that $f$ has a fixed point and since $f$ is strictly decreasing, this fixed point is unique. Since this fixed point is
also a fixed point of $f^{2n+1}$, the fact that $f^{2n+1}$ is strictly decreasing shows that the unique fixed point of $f^{2n+1}$ is the same as the
fixed point of $f$, and in particular must have period 1.
Since $f^2$ is strictly increasing, the periodic points of $f^2$ must have period 1. This shows that any fixed point of $f^{2n}$ must also be a fixed
point of $f^2$.
Hence we have $\operatorname{Per}(f) = \{ x | f(x) =x \} \cup \{ x | f^2(x) = x \}$.
