We can modify the connectedness argument to show it also holds for homeomorphisms: suppose $\omega(x)$ is contained in a ball $B(0,R)$, with $R>0$, and let $N$ be so big that $T(B(0,2R))\subseteq B(0,N)$.
Then if the positive orbit of $x$ is not bounded, it must contain infinite points inside $B(0,N)\setminus B(0,2R)$, which contradicts the fact that $\omega(x)\subseteq B(0,R)$.