Let $X$ be a set. For $B\subseteq X$ and ${\cal H}\subseteq {\cal P}(X)$ we set $$\text{ST}^1(B,{\cal H}) = \{H\in {\cal H}: H\cap B\neq \emptyset\},$$ and $\text{st}^1(B,{\cal H}) = \bigcup \text{ST}^1(B,{\cal H})$. For any integer $n>1$ we inductively set $$\text{ST}^{n+1}(B,{\cal H}) = \{H\in {\cal H}: H\cap\text{st}^n(B,{\cal H}) \neq \emptyset\},$$ and $\text{st}^{n+1}(B,{\cal H}) = \bigcup \text{ST}^{n+1}(B,{\cal H})$.

Finally, a topological space $(X,\tau)$ is said to be *$n$-star-compact* if for every open cover ${\cal U}$ of $X$ there is a finite subset ${\cal V}\subseteq {\cal U}$ such that $$\text{st}^n(\bigcup{\cal V},{\cal U}) = X.$$ Obviously, any $n$-star-compact space is $(n+1)$-star-compact.

Given any $n\geq 1$, what is an example of a space that is $(n+1)$-star-compact, but not $n$-star-compact?

(...) there is a finite $\mathcal{V}\subset\mathcal{U}$" ? (I mean, not necessarily a subcover, otherwise the subsequent condition becomes trivial) $\endgroup$ – Pietro Majer May 29 '18 at 18:38