The hard part of this is figuring out what $\omega(t,\sigma)$ looks like. I’ll do that and leave the rest to you for now.
Let $x\in\Sigma_2$ be any sequence with at least two ones that is not in the orbit of $t$. Then there are $k,\ell\in\Bbb Z^+$ such that $k<\ell$, $x_k=x_\ell=1$, $x_n=0$ for $k<n<\ell$, and either
- $x_{2\ell-k+1}\ne 1$, or
- $x_n=1$ for some $n$ such that $\ell<n<2\ell-k+1$.
In other words, $x$ has a substring
$$1,\underbrace{0,0,\ldots,0,0}_{\ell-k-1\text{ zeroes}},1$$
that is not immediately followed by
$$\underbrace{0,0,\ldots,0,0}_{\ell-k\text{ zeroes}},1\;,$$
as it would be in any member of the orbit of $t$. Then
$$\{y\in\Sigma_2:y_n=x_n\text{ for }k\le n\le2\ell-k+1\}$$
is an open nbhd of $x$ disjoint from the orbit of $t$. On the other hand, it’s straightforward to verify that every $x\in\Sigma_2$ with at most one $1$ is in $\omega(t,\sigma)$, so
$$\omega(t,\sigma)=\left\{\sigma^n(t):n\ge 0\right\}\cup\{z\}\cup\{e^{(n)}:n\in\Bbb Z^+\}\;,$$
where $z$ is the zero sequence, and $e^{(n)}=\langle\delta_{nk}:k\in\Bbb Z^+\rangle$ is the sequence whose $n$-th term is $1$, all other terms being $0$.