Metric spaces and σ-closure-preserving bases (Nagata's metrization theorem) 
The first two line of the proof say that if $X$ is metrizable (so paracompact) then clearly there is a base $\mathcal{G}= \bigcup_{i\in \mathbb{N}} G_i$ having this property. Paracompactness just gives locally finite open refinement for any open cover of $X$. So could you please tell me how I find such a base?
(The image above is from Harold W. Martin, A note on the Frink metrization theorem, Rocky Mountain J. Math., Vol.6, No.1 (1976), pp.155-157, doi:10.1216/RMJ-1976-6-1-155.)
 A: For each $n\in\Bbb Z^+$ let $\mathscr{G}_n$ be a locally finite open refinement of the open cover 
$$\left\{B\left(x,\frac1n\right):x\in X\right\}\;.$$
Locally finite families are closure-preserving, so each $\mathscr{G}_n$ is closure preserving. Let $x\in X$ and $n\in\Bbb Z^+$; then $\{G\in\mathscr{G}_n:x\in G\}$ is finite, so $V_n(x)=\bigcap\{G\in\mathscr{G}_n:x\in G\}$ is an open nbhd of $x$. Fix any $G_x\in\{G\in\mathscr{G}_n:x\in G\}$; then $G\subseteq B\left(y,\frac1n\right)$ for some $y\in X$, so 
$$\operatorname{diam}V_n(x)\le\operatorname{diam}G_x\le\operatorname{diam}B\left(y,\frac1n\right)\le\frac2n\;.\tag{1}$$
Thus, for each $n\in\Bbb Z^+$ we have 
$$x\in V_{3n}(x)\subseteq B\left(x,\frac1n\right)\;,$$
and $\{V_n(x):n\in\Bbb Z^+\}$ is therefore a local base at $x$.
Added: Let $\mathscr{G}=\bigcup_{n\in\Bbb Z^+}\mathscr{G}_n$. The fact that $\mathscr{G}$ is a base for $X$ also follows easily from $(1)$: for any $G\in\mathscr{G}_{3n}$ we have
$$x\in G\subseteq B\left(x,\frac1n\right)\;.$$
