Expansion lemma for paracompact hausdorff space. 
[Munkres, Section41, Ex5] Let $X$ be paracompact. We proved a "shrinking lemma" for arbitrary indexed open coverings of $X$. Here is
  an "expansion lemma" for arbitrary locally finite indexed families in
  $X$.
Lemma. Let $\{ B_ \alpha \} _ { \alpha \in J}$ be a locally finite
  indexed family of subsets of the paracompact Hausdorff space $X$. Then
  there is a locally finite indexed family $\{ U_ \alpha \} _ { \alpha
 \in J}$ of open sets in $X$ such that $ B_ \alpha \subset U_ \alpha$
  for each $\alpha$.

In the proof of the shrinking lemma(Munkres lemma41.6), the author uses the collection of all open sets such that the closure is contained in the given set. So I used the collection of all open sets containing $B_ \alpha$ for some $ \alpha$. But this seems to not work. How can I prove this?
 A: For each $x\in X$ let $V(x)$ be an open nbhd of $x$ that meets only finitely many members of $\mathscr{B}$. Let $\mathscr{W}$ be a locally finite open star refinement of $\{V(x):x\in X\}$. 
For each $\alpha\in J$ let $U_\alpha=\bigcup\{W\in\mathscr{W}:W\cap B_\alpha\ne\varnothing\}$; clearly $B_\alpha\subseteq U_\alpha$, and $U_\alpha$ is open. Let $\mathscr{U}=\{U_\alpha:\alpha\in J\}$.
Let $W_0\in\mathscr{W}$ be arbitrary; there is a $y\in X$ such that $$\operatorname{st}(W_0,\mathscr{W})=\bigcup\{W\in\mathscr{W}:W\cap W_0\ne\varnothing\}\subseteq V(y)\;.$$ Thus, if $F_0=\{\alpha\in J:B_\alpha\cap\operatorname{st}(W_0,\mathscr{W})\ne\varnothing\}$, then $F_0$ is finite. Suppose that $W_0\cap U_\alpha\ne\varnothing$ for some $\alpha\in J$; then there is a $W\in\mathscr{W}$ such that $W\cap B_\alpha\ne\varnothing\ne W\cap W_0$, so $B_\alpha\cap\operatorname{st}(W_0,\mathscr{W})\ne\varnothing$, and $\alpha\in F_0$. That is, each $W\in\mathscr{W}$ meets only finitely many $U\in\mathscr{U}$.
Now let $x\in X$ be arbitrary; $\mathscr{W}$ is locally finite, so $x$ has an open nbhd $G$ that meets only finitely many members of $\mathscr{W}$. We’ve just seen that each $W\in\mathscr{W}$ meets only finitely many members of $\mathscr{U}$, so $G$ meets only finitely many members of $\mathscr{U}$. Since $x$ was arbitrary, $\mathscr{U}$ is locally finite, as desired.
