The closure relationship of a set in two typologies Space ‎$ ‎(X,‎\tau ‎)‎ $‎ is a space that  every open cover has a finite subfamily whose closures cover the space. ‎$ A‎ ‎‎\subseteq ‎X‎ $‎ also has this property,in other words every open cover  of $\tau$ for $A$ has a finite subfamily whose closures cover $A$.
Now I want to show that space  ‎$ (X , ‎\tau(X\setminus A)‎ $‎‎ also has this property. For this purpose, I put $ X = A ‎\cup (‎ X‎‎‎\setminus ‎A)‎ $‎, and I want to show that sub-sets $A$ and $‎‎X\setminus A$  each have this property with topology $\tau (X\setminus A )$( the simple extension topology $\tau$ by $A$), so space $( X= A‎\cup (‎X‎\setminus ‎A),‎\tau(X‎‎\setminus A))$ also has this property.
I have no problem with part $‎X\setminus A $, but I had a problem with part $A$. Please help me.

My efforts:

Let $\tau(X\setminus A) =\tau_{1}$ and ‎‎‎$ ‎‎‎U =‎‎\{ ‎u‎_i=‎v_i‎‎‎‎‎\cup(‎‎w_i‎‎\setminus ‎A):i\in I‎‎‎‎ \} $ ‎be a‎ ‎‎$ \tau_1 $- ‎open ‎cover ‎‎$‎X‎$‎.(‎$‎v_i,w_i\in\tau ‎$‎).We Know ‎$\{v_i :‎i‎‎\in I‎ \}$ ‎is a ‎‎‎‎$‎\tau‎$‎-cover ‎for ‎‎$‎A‎$‎.
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According to ‎property‎ ‎$‎A‎$‎, there is a finite subset ‎$   ‎I_0  ‎‎\subseteq ‎I‎  $‎‎so that ‎$ A‎ ‎‎\subseteq  ‎‎‎\bigcup‎_{i \in I_0}‎‎‎ ‎\overline{v_i}‎ $‎.
Also, due to the feature of space ‎‎$‎(X, \tau‎)‎$‎‎, there ‎is‎ finite ‎subset $   ‎I_{1}‎  ‎‎\subseteq ‎I‎ ‎$‎‎, which ‎$ ‎X‎‎ ‎‎\subseteq  \displaystyle\mathop{\bigcup}_{i\in ‎‎I_{1}}\hspace{-.9mm}‎‎‎ ‎\overline{v_i ‎\cup ‎w_i‎‎}‎ $‎‎‎. we ‎put‎‎$ J‎ =‎ ‎I_1‎ ‎\cup ‎I_0‎ $‎.‎ ‎
In ‎topology ‎‎$‎\tau‎$ ,‎We got ‎to‎ $\displaystyle\mathop{\bigcup}_{i\in ‎J}\hspace{-.9mm} ‎‎‎‎\overline{‎u‎_i‎‎} \supseteq X‎ ‎‎\setminus ‎A‎ ‎$
‎But how can I prove the ‎phrase$(\displaystyle\mathop{\bigcup}_{i\in ‎J}\hspace{-.9mm} ‎‎‎‎\overline{‎u‎_i‎‎} )_{\tau_1} \supseteq X‎ ‎‎\setminus ‎A‎ ‎$‎?‎‎‎
‎Note

In topological space $(X,\tau(A)$ and $(X,\tau)$ and $B \subset X$ we have the below relation:

‎$ ‎\overline{B}‎‎‎_{‎\tau(A)‎} =\overline{B}‎_{‎\tau‎‎}‎ ‎\cap ‎(A‎^{c}‎‎ ‎\cup (‎ A‎ ‎‎\cap‎‎ ‎\overline{( B ‎\cap A‎})‎_{\tau}‎‎ ‎‎‎‎‎$‎‎‎
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 A: Given a topology $\tau$ on $X$, we form $\tau_A$ as the smallest topology that contains $\tau$ and has $A$ as a closed subset, i.e. $X\setminus A \in \tau_A$. This has open sets that are either in $\tau$ or are of the form $O \setminus A$, where $O \in \tau$.
Now, if $\mathcal{U}=\{U_i\mid U_i \in \tau, i \in I\} \cup \{O_j\setminus A\mid O_j \in \tau, j \in J\}$ is an open cover of $X$ wrt $\tau_A$, then $\{U_i\mid i \in I\}$ is an open cover of $A$ wrt $\tau$, so by assumption there are finitely many $U_i, i \in F_I$ (so $F_I \subseteq I$ finite) such that their closures (in $\tau$) cover $A$. Of course the collection $\{U_i, O_j \mid i \in I, j \in J\}$ is a $\tau$-open cover of $X$, so here we also have a finite $F'_I, F_J$ (finite subsets of $I$ resp. $J$), such that the $\tau$-closures of the corresponding $U_i$ and $O_j$ covers $X$.
Now let $x \in X$. If $x \in A$, an open neighbourhood of $x$ is just some open $O \in \tau$ that contains $x$. There is some $i \in F_I$ such that $x$ is in the $\tau$-closure of $U_i$. This $O$ thus intersects $U_i$ for that $i$ and $x$ is in the $\tau_A$ closure of $U_i$. If $x \notin A$, For some $i \in F'_I$ or $j \in F_J$, $x$ is in the $\tau$-closure of $U_i$ resp $O_j$. A (basic) open neighbourhood of $x$ is of the form $O\setminus A$, where $O \in \tau$, and we know $O$ intersects either $U_i$ or $O_j$, so $O \setminus A$ intersects $U_i$ or $O_j \setminus A$. So $x$ is in the $\tau_A$-closure of the same set as it was before wrt $\tau$ as well. So $(X,\tau_A)$ has the same "pseudo-"-H-closed property as $X$ has.
