question about the closures of a set A ‎subset ‎$‎‎A‎$ ‎(‎resp.‎ ‎subspace ‎‎$‎A‎$‎) ‎of ‎‎space ‎$  ‎(‎‎X‎,\tau)$ ‎has ‎‎$‎\alpha‎$‎-‎property ‎if  ‎every ‎$‎\tau‎$‎-open ‎(‎resp.‎‎‎$‎\tau‎_{‎A‎}‎‎$‎-‎open) ‎‎‎cover of ‎‎$‎A‎$‎ has a finite subfamily whose ‎‎$‎\tau‎$‎‎-closures ‎(resp.‎‎ $‎\tau‎‎_{‎A‎}‎$‎‎-‎closures)‎ ‎cover ‎‎$‎A‎$‎.‎
If subspace $A$ ‎has ‎‎$‎\alpha‎$‎-‎property of topological space $(X, \tau)$, does closures $A$ have $‎\alpha‎$‎-‎property as a subspace?
 A: Notation. Let $A \subseteq X$ be any subset. For convenience, we denote the $\tau_{A}$-closure of a subset  $U$ of $A$ as $\mathrm{cl}_{A}(U)$, and the $\tau_{X}$-closure of $A$ as $\overline{A}$.
Little exercises for you.
(a) $\mathrm{cl}_{A}(U)= \overline{U} \cap A$. (Hint: Use definitions of subspace topology and closure.)
(b) $\overline{U_{1}\cup U_{2}} =\overline{U_{1}}\cup \overline{U_{2}} .$
With these preparations, we are able to prove the proposition. Suppose $A$ has $\alpha$-property. For the purpose of showing that $\overline{A}$ also has $\alpha$-property, we take any $\tau_{\overline{A}}$-open cover $\{U_{i} \cap \overline{A}\}_{i \in I}$ of $\overline{A}$. Needless to say, $\{U_{i} \cap A\}_{i \in I}$ is a $\tau_{A}$-open cover of $A$. Since $A$ has $\alpha$-property, there exists a finitely many indices $i_{1},i_{2},\dots,i_{n}$ such that $\bigcup_{k=1}^{n} \mathrm{cl}_{A}(U_{i_{k}}\cap A) = A$. According to exercises above, this means $$\bigcup_{k=1}^{n}\overline{U_{i_{k}} \cap A} \supseteq A.$$
Notice the set on the left side, as a finite union of closed subsets, is closed. We actually have $$\bigcup_{k=1}^{n}\overline{U_{i_{k}} \cap A} \supseteq \overline{A}.$$
Moreover this implies $$\bigcup_{k=1}^{n}\overline{U_{i_{k}} \cap \overline{A}} \supseteq \overline{A}.$$
The upshot is that now we obtain $\bigcup_{k=1}^{n} \mathrm{cl}_{\overline{A}}(U_{i_{k}} \cap \overline{A}) = \overline{A}$, by exercise (a) again, which means $\overline{A}$ also has $\alpha$-property.
