Let $f\colon X\to X$ be a continuous map on a (compact) topological space $X$. Let $A$ be an attractor for $f$, i.e. there is a neighborhood $V$ of $A$ such that $f^N(V)\subset V$ for some $N\in\mathbb{N}$ and $A=\bigcap_{n\in\mathbb{N}}f^n(V)$. Now for $x\in V$, consider the set $\omega(x)$ of all $\omega$-limit points of $x$, i.e. $$ \omega(x)=\bigcap_{n\in\mathbb{N}}\overline{\left\{f^k(x): k>n\right\}}. $$ (You can see this definition of $\omega(x)$ here if you like.)
Now, I am interested to know whether $$ \omega(x)\subset A~\forall x\in V. $$
Let $x\in V, y\in\omega(x)$. Then, by definition, $$ y\in\overline{\left\{f^k(x): k>n\right\}}~\forall n\in\mathbb{N}.~~~~~(*) $$ In order to have $y\in A$, it would be necessary that $$ y\in f^n(V)~\forall n\in\mathbb{N}.~~~~~~~(+) $$
I do not see that in general $(*)$ does imply $(+)$.
Edit
If $V$ is closed (I guess this is the case since it is not said that $V$ is an open neighborhood of $A$) and $f(V)\subset V$, i.e. $N=1$, then $(*)$ should imply that $y\in f^n(\overline{V})=f^n(V)$ for all $n\in\mathbb{N}$, i.e. in fact $y\in A$.
But, as mentioned, I do not see that in general. Maybe you do?