property $\beta$ 
A topological space $‎(X,‎\tau‎  )‎$ has property *  ‎‎, if every open cover of ‎‎$X‎$‎ has a finite subfamily whose closures cover $X$.


A topological space $‎(X,‎\tau‎  )‎$ has property $\beta$  ‎‎, if every closed  set of ‎‎$X‎$‎  according to $\tau$ has property * .


Question: Let $‎(X,‎\tau‎  )‎$ and $A \subset X$  have property $\beta$. Does $(X, \tau(X- A))$ have property $\beta$?


$(X, \tau(X- A))$ is topological space of  the simple expansion of ‎‎τ‎‎ by $( ‎X−‎A‎‎ )$

 A: I guess the answer is no if I correctly understood the definition of simple extension. I understood the topology $\tau(X-A)$ as one generated by the sub-basis $\tau \cup \left\{ X-A \right\}$. According to this definition, if $A$ were closed, then nothing would change.
Let $X = \mathbb{R}$, the usual real line and $A=\emptyset$. Then $A$ is $\beta$-space vacuously. Moreover $Y=(X,\tau(X-A))$ is nothing but $(X,\tau)$ itself. Now consider the open cover $\mathcal{U} = \{[n,n+1]|n\in \mathbb{Z}\}$ of $Y$. For it to be a $\beta$-space, $Y$ itself should have the $\ast$ property. However, no closure of any finite subset of $\mathcal{U}$ can cover $Y$ since it would be a bounded set.
As seen in the above, the original question is somewhat ill-posed. However, I guess a slight modification will produce a good exercise.

Let $X$ be a topological space with the topology $\tau$ and consider a new topology generated by
$$\tau \cup \left\{ X \setminus A| \textrm{$A$ is $\ast$ }\right\}$$. Determine if one can say that $X$ is $\ast$ space.

