Let $X$ is a linearly lindelof subspace of $Z$ and $b$ is not $\omega$separated from $X$, i.e., for any closed $G_\delta$ set $P$ of $Z$ which contains $b$, $P\cap X \not=\emptyset$. If $\tau < \aleph_\omega$, how to show that $b$ is not $\tau$separated from $X$, i.e., for any closed $G_\tau$ set $P$ of $Z$ which contains $b$, $P\cap X \not=\emptyset$? Thanks ahead:)

I think you need some separation axioms (see my comment above). Here is a proof when $Z$ is regular. I hope I did not overlook something. Suppose that there is some closed $G_\tau$ $P\ni b$ disjoint from $X$, so $P=\cap_{\alpha<\tau}U_\alpha$ for $U_\alpha$ open in $Z$. By regularity of $Z$, one can choose inductively open subsets $V_{\alpha,n}\ni b$ such that $\overline{V_{\alpha,n+1}}\subset V_{\alpha,n}$, with $V_{\alpha,0}=U_\alpha$. Then $b\in Q=\cap_{\alpha<\tau}\cap_{n<\omega}V_{\alpha,n}= \cap_{\alpha<\tau}\cap_{n<\omega}\overline{V_{\alpha,n}}\subset P$, so $Q$ is also a closed $G_\tau$ set disjoint from $X$. Thus, for each $x\in X$ there is $\alpha,n$ such that $x\not\in\overline{V_{\alpha,n}}$. This yields that $\mathcal{W}=\{ Z\overline{V_{\alpha,n}}:\alpha\in\tau,n\in\omega\}$ is a cover of $X$. It is easy to see that a space is linearly Lindelöf iff any cover of uncountable cofinality has a subcover of countable cofinality. Since $\omega$ is the only cardinal of countable cofinality smaller than $\tau$, $\mathcal{W}$ has a countable subcover. Looking at complements again, this means that the intersection of the corresponding $V_{\alpha,n}$ is disjoint from $X$. This gives a $G_\delta$ set containing $b$ which is disjoint from $X$. To obtain a closed one (and thus a contradiction), do again the trick of taking a nested intersection inside each element, as when defining the $V_{\alpha,n}$. 

