Let $h$ be a homeomorphism of $\mathbb{R}^2$ onto itself such that $h(K)=K$ for some compact subset $K$ of $\mathbb{R}^2$. Show that $h$ has a periodic point in $\mathbb{R}^2$.
My idea is consider a sequence $x, f(x),ff(x),...$ for some $x\in K$, since $K$ is compact this sequence has a cluster point at $K$. But I couldn't go further.
Use the second Brouwer plane transformation theorem listed in your other post. If the map were fixed point free, then we could transform it by a homeomorphisminto a map of the plane that is just translation; however, the image of the compact set would be compact, and translation never preserves a compact set (it could preserve an infinite line, but not a compact set; what would the image of the points with largest $x$-coordinate be?)
More specifically, look at $\phi^{-1}(K)$, where $\phi$ is the map discussed in Brouwer transformation plane theorem. This set is compact, so it is bounded, so eventually translation does not preserve it, in fact there is an $n$ such that $\tau^n (\phi^{-1}(K)))\cap \phi^{-1}(K)=\emptyset$. But $\phi\tau^n=h^n\phi$, so taking $\phi$ of the two sets in the last equation, we see that $h^n(K)\cap K=\emptyset$.