Closure preserving A metrizable space X has an open basis $\mathcal{G}$ which represented as a sequence $G_1 , G_2, G_3 ...$
of closure preserving collections.
Then since  every metric space is paracompact, so clearly there exist a locally finite open refiniment $\mathcal{H}$  for $\mathcal{G}$   which is also closure preserving collections. How??
$\bullet  G_n$ is said to be closure preserving if 
$\overline{\cup A} = \cup \overline{A}$ forall $A\in G^*$ where $G^*$ is subfamily of $G_n$.
$\bullet$ $\mathcal{H}$ is refinement of $\mathcal{G}$ means for every $H\in \mathcal{H}$ there is a $G \in \mathcal{G}$ s.t $H \subseteq G$.    
I was thinking to show $\mathcal{H}= \cup_{i\in \mathbb{N}}{H_n}$, where $H_n$ is closure preserving and assuming for every $A_i \in H_n ~\exists ~B_i \in G_n$ s.t $A_i \subseteq B_i$
 A: The trick to getting local finiteness of $\mathscr{H}$ is to do it in two steps.
Let $\mathscr{G}=\bigcup_{n\in\Bbb Z^+}\mathscr{G}_n$ be the given base, so that each $\mathscr{G}_n$ is closure-preserving. For $n\in\Bbb Z^+$ let $U_n=\bigcup\mathscr{G}_n$; then $\mathscr{U}=\{U_n:n\in\Bbb Z^+\}$ is an open cover of $X$ and therefore has a locally finite open refinement $\mathscr{R}$. For each $R\in\mathscr{R}$ there is an $m(R)\in\Bbb Z^+$ such that $R\subseteq U_{m(R)}$. For each $n\in\Bbb Z^+$ let $V_n=\bigcup\{R\in\mathscr{R}:m(R)=n\}$; then $\mathscr{V}=\{V_n:n\in\Bbb Z^+\}$ is a locally finite open refinement of $\mathscr{U}$ such that $V_n\subseteq U_n$ for each $n\in\Bbb Z^+$.
Let $n\in\Bbb Z^+$; clearly $\{G\cap V_n:G\in\mathscr{G}_n\}$ is an open cover of $V_n$. $X$ is metrizable, and $V_n$ is a subspace of $X$, so $V_n$ is metrizable and hence paracompact. Thus, we can let $\mathscr{H}_n$ be a locally finite open refinement of $\{G\cap V_n:G\in\mathscr{G}_n\}$ covering $V_n$. (Technically the members of $\mathscr{H}_n$ are open in $V_n$, but since $V_n$ is open in $X$, $\mathscr{H}_n$ is actually a family of open sets in $X$.)
Now let $\mathscr{H}=\bigcup_{n\in\Bbb Z^+}\mathscr{H}_n$; clearly $\mathscr{H}$ is an open refinement of $\mathscr{G}$ covering $X$, and you should have little trouble showing that $\mathscr{H}$ is locally finite.
To complete the argument, just show that every locally finite family of sets is closure preserving. This is very straightforward; it uses the fact that if $\mathscr{F}$ is any finite family of sets, then
$$\operatorname{cl}\bigcup\mathscr{F}=\bigcup_{F\in\mathscr{F}}\operatorname{cl}F\;.$$
