Eventually periodic point and homeomorphism. The point $x$ is a periodic point of period $n$ if $f^n(x)=x$. The least positive $n$ for which $f^n(x)=x$ is called the prime period of $x$.
A point $x$ is eventually periodic of period $n$ if $x$ is not periodic but there exists $m>0$ such that $f^{n+i}(x)=f^i (x)$ for all $i\geq m$.
First. Prove that a homeomorphism of R can have no periodic point points with prime period greater than $2$.
Second. Prove that a homeomorphism cannot have eventually periodic points.
Any suggestions, please? 
 A: Since you tagged this calculus rather than general topology, I’ll answer in calculus terms.
If $f:\Bbb R\to\Bbb R$ is a homeomorphism, it must be both continuous and injective. Use that to prove that $f$ must be a strictly monotone function; the intermediate value theorem is useful here. (It can be either increasing or decreasing.) 
Next, show that if $f$ is strictly increasing, and $x\in\Bbb R$, then either $f(x)=x$, or the points $f^n(x)$ for $n\ge 0$ are all distinct. (If $x<f(x)$, what can you say about the sequence $\langle x,f(x),f^2(x),\dots\rangle$? What if $x>f(x)$?)
Finally, show that if $f$ is strictly decreasing, then $f\circ f$ is strictly increasing, and conclude that $f$ can have only fixed points, points of prime order $2$, and points whose orbits are infinite. (The orbit of $x$ is the collection of points $f^n(x)$.)
Added: Although the line of argument sketched above provides a bit more information, you can (as Thomas Andrews mentioned in the comments) show that an injection cannot have any eventually periodic points. Let $f:X\to X$ be injective, and suppose that $x$ is eventually periodic with prime period $n$ and first period starting with $f^m(x)$. Let $y=f^{m-1}(x)$ and $z=f^{m+n-1}(x)$; then $y\ne z$, but $f(y)=f^m(x)=f(z)$, contradicting the injectivity of $f$.
