 Solution

A compact metric space is second countable. To see this concretely, for each positive integer $r$ choose a finite cover by balls of radius $1/r$. Their centres, over all $r$, form a countable dense subset; balls centred there with positive rational radii form a countable basis of a topology. Enumerate its nonempty members as $(V_j{)}_{j\ge 1}$, repeating members if the basis is finite.

For each $j$, putEach $G_j$ is an open set, because all iterations of a map $f^n$ are continuous. It is dense: if $U$ is any nonempty open set, the hypothesis supplies $n\ge 1$ and $u\in U$ with $f^n(u)\in V_j$, so $u\in U\cap G_j$.

A compact metric space is a complete metric space, hence a Baire space. The Baire category theorem implies that $G={\bigcap }_{j\ge 1}{G}_j$ is dense, and in particular nonempty. For any $x\in G$, the forward orbit of $x$ meets every $V_j$, and therefore every nonempty open set. Its orbit closure is $X$. Thus the open-set intersection hypothesis implies point transitivity, and in fact the set of points with a dense forward orbit contains the dense intersection $G$.