Vtyjkyuk

Let $f: X rightarrow X$ be a homeomorphism of a compact, connected metric space. If the orbit of $x$ is compact, then $x$ is periodic.

I feel like should be trivial, but I cannot seem to work out a proof.

Since $f(Orb(x)) = Orb(x)$ and $Orb(x)$ is minimal, it follows that $Orb(x)$ is nowhere dense.

Assuming that the orbits are two-sided orbits consider the orbit $Y$ of $x$. This is a countable compact metric space and $f$ is a homeomorphism of $Y$ onto itself. Write $Y$ as the union of its singletons and apply BCT to conclude that some point in $Y$ is open. This implies that each point of $Y$ is open because $f$ is a homeomorphism. Compactness of $Y$ implies that $Y$ is a finite set. [Look at the open cover formed by singletons].

Kavi Rama Murthy's proof can be modified to work when the orbit $O=x_0,f(x_0),f(f(x_0)),dots$ is one sided. As in their proof, some point $x$ in $O$ is open in $O$. This means there is an open $Usubset X$ for which $Ucap O=x$. This implies $f(x)$ is open as well, as
$$
f(x)=f(Ucap O)=f(U) cap f(O)=(f(U)setminus x_0)cap O
$$
Note $f(Acap B)=f(A)cap f(B)$ since $f$ is injective.

Supposing $x$ is not the initial point $x_0$, then

beginalign
f^-1(x)
&=f^-1(Ucap O)
\&=f^-1(U)cap f^-1(O)
\&=f^-1(U)cap (Ocup f^-1(x_0))
\&=(f^-1(U)cap O)cup f^-1(Ucap x_0)
\&=f^-1(U)cap O
endalign
The last equality follows since $Ucap O=x$, so $x_0notin U$. This shows $f^-1(x)$ is open.

Therefore, the fact $x$ is open implies all points in the orbit are open, and you can conclude as in the other proof.