Proposition. Let $(X,d)$ a metric space and let $A\subseteq X$ a subset of $X$ such that $A$ is separable (then $A$ has a dense -with respect the subspace topology of $A$ - and countable subset $D$). Let $W\subseteq X$ an open subset of $X$. Let $\mathbb{B}alls(X)$ be the family of all closed balls of $X$. If $\mathcal{H}_1,\dots,\mathcal{H}_z$ are subfamilies of $\mathbb{B}alls(X)$ such that the family $\mathcal{H}_j$ is disjointed for all $j=1,\dots, z$ and $H\subseteq W$ for all $H\in\mathcal{H}_j$ and for all $j=1,\dots,z$, and if
$A\cap W\subseteq\bigcup_{j=1}^z\bigcup_{H\in\mathcal{H}_j}H$,
then for each $j=1,\dots,z$ there exists a countable subfamily $\mathcal{G}_j$ such that
$A\cap W\subseteq\bigcup_{j=1}^z\bigcup_{G\in\mathcal{G}_j}G$.
I have got difficulties to give a proof of the above proposition. Can anyone help me, please?
My idea is this one but I don't know if it works. First of all I observe that $A\cap W$ is open in $A$ (with respect to the subspace topology of $A$). Then $A\cap W$ is an open subset of the separable space $A$, hence it's a separable space. Then there exists a countable subsets $Q$ of $A\cap W$ such that
$Cl_{A\cap W}(Q)=A\cap W$
where $Cl_{A\cap W}$ denotes the closure with respect the subspace topology of $A\cap W$. By hypothesis
$Q\subseteq\bigcup_{j=1}^z\bigcup_{H\in\mathcal{H}_j}(H\cap A\cap W)$.
Since $Q$ is countable, we deduce that for each $j=1,\dots,z$ there exists a countable subfamily $\mathcal{G}_j$ such that
$Q\subseteq\bigcup_{j=1}^z\bigcup_{G\in\mathcal{G}_j}(G\cap A\cap W)=\bigcup_{j=1}^z\bigcup_{G\in\mathcal{G}_j}(G\cap A\cap W)$.
Taking the closure of both sides we obtain that
$A\cap W\subseteq Cl_{A\cap W}\left(\bigcup_{j=1}^z\bigcup_{G\in\mathcal{G}_j}(G\cap A) \right)= \bigcup_{j=1}^z Cl_{A\cap W}\left(\bigcup_{G\in\mathcal{G}_j}(G\cap A)\right) =\bigcup_{j=1}^z\bigcup_{G\in\mathcal{G}_j}Cl_{A\cap W}(G\cap A)=\bigcup_{j=1}^z\bigcup_{G\in\mathcal{G}_j} G\cap A$.
I am sure the each ot the above equalities works eccept for the second-last one, that is to say
$\bigcup_{j=1}^zCl_{A\cap W}\left(\bigcup_{G\in\mathcal{G}_j}(G\cap A)\right)= \bigcup_{j=1}^z\bigcup_{G\in\mathcal{G}_j}Cl_{A\cap W}(G\cap A)$.
I think that it is true but I don't how to show it. My idea is to show (BUT I DON'T KNOW HOW TO DO IT) that the family
$\mathcal{F}:=$ {$G\cap A\mid G\in\mathcal{G}_j\text{ for some }j=1,\dots, z$ }
is a locally finite family of subsets of $A\cap W$, that is to say that for each $a\in A\cap W$ there exists an open subset $U$ of $a$ such that the elements $F\in\mathcal{F}$ such that $U\cap F\ne\emptyset$ is finite.