question about finer topology ‎‎A topological space $X$ is said to be star compact if whenever $\mathscr{U}$ is an open cover of $X$, there is a compact subspace $K$ of $X$ such that $X = \mathrm{st}(K,\mathscr{U}),$ where
$\mathrm{st}(K, \mathscr{U})
= \bigcup \{ U \in \mathscr{U}: U \cap K \neq \emptyset \}.$
We recursively define $\mathrm{st}^n$ for $n=0,1,2,\ldots$ by
$$\begin{align}
\mathrm{st}^0(K, \mathscr{U})
&= K
\\
\mathrm{st}^{n+1}(K, \mathscr{U}‎)
&= \bigcup ‎\{ U‎ ‎\in‎ \mathscr{‎U} : U ‎\cap\  \mathrm{st}^n(K, \mathscr{U}‎) \neq ‎\emptyset \}‎‎‎
\end{align}$$
Definition: ‎A space $X$ is said to be ‎$ ‎\omega‎$‎-starcompact if for every open cover ‎$\mathscr{U}‎$‎ of ‎‎$X$‎, there is some ‎‎$‎n ‎\in ‎\mathbb{N‎}‎‎^{+}‎$‎ and some finite subset ‎‎$B$‎ of ‎‎$X$‎ such that ‎‎$\mathrm{st}^{‎n‎}(B, \mathscr{U}) = X‎$‎.‎‎
‎

Let‎‎ ‎$(X, ‎\tau)$ be‎ $\omega‎$‎-‎starcompact and $\tau^{*} \subset \tau$. Is $(X, \tau^{*})$ a‎ $‎\omega‎ $‎-‎starcompact ‎space?‎


Yes, I think so!  
To see this, fix $\mathscr{U} \subseteq \tau^* \subseteq \tau $, an open cover of $X$, and pick a finite set $B \subseteq X$ and whole number $n$ for which $st^n(B,\mathscr{U}) = X$.  
Note $st(B,\mathscr{U})$ is the same open set (i.e. the same union of open sets from the open cover) regardless of which topology we consider.  In fact, for all $k$, $st^k(B,\mathscr{U})$ is the same union of open sets from the open cover regardless of which topology we consider.  As such, $st^n(B,\mathscr{U}) = X$, regardless of which topology we consider.
