 Solution

Let $x$ have a dense forward orbit $O=\{f^j(x):j\ge 0\}$, and suppose that $f$ is surjective. Every tail $O_k=\{f^j(x):j\ge k\}$ is then dense. Indeed, continuity givesand the left side equals $f^k(X)=X$ by surjectivity.

Given nonempty open sets $U,V$, choose $k\ge 0$ with $f^k(x)\in U$. The dense tail $O_{k+1}$ supplies $j\ge k+1$ with $f^j(x)\in V$. Taking $n=j-k\ge 1$ gives $f^j(x)\in f^n(U)\cap V$. This proves the required positive-time intersection property.

The result can fail without surjectivity. Take the compact metric spacewith the metric induced from the real line. All nonzero points are isolated, so continuity there is automatic; at zero, $f(1/j)\to 0=f(0)$ proves continuity. The forward orbit of $1$ is $\{1,1/2,1/3,\dots \}$, whose closure is $X$, so this is point transitivity. But $f(X)=X\setminus \{1\}$. The singletons $U=\{1/2\}$ and $V=\{1\}$ are nonempty open sets, and for every $n\ge 1$,The isolated initial point can be visited only at the start of the dense forward orbit; deleting that initial segment destroys density.