Motivation for a proof "In a regular space, if every open cover contains a countably locally finite open refinement, then the space is paracompact". Let $(X,\tau)$ be a regular topological space.
It's a theorem that the followings are equivalent:
(1) Every open cover has a countably locally finite open refinement.
(2) Every open cover has a locally finite refinement.
(3) Every open cover has a locally finite closed refinement.
(4) Every open cover has a locally finite open refinement.
An argument i know proves the theorem via $(1)\Rightarrow (2) \Rightarrow (3) \Rightarrow (4) \Rightarrow (1)$.
From (1) to (3), an argument is very clear to me.
However, i don't really get the motivation of the proof for (3) $\Rightarrow$ (4).
The argument is as follows:
Let $\mathcal{A}$ be an open cover of $X$.
Then, there exists a locally finite closed refinement $\mathcal{B}$ of $\mathcal{A}$.
Now, define $G\triangleq \{V\in\tau: \{B\in\mathcal{B}|B\cap V \neq \emptyset\} \text{ is finite} \}$.
It's clear that $G$ is an open cover of $X$.
Now, take $\mathcal{C}$ as a locally finite closed refinement of $G$.
Next, define $S(B)=\{C\in\mathcal{C}:C\cap B = \emptyset \}, \forall B\in\mathcal{B}$.
And, let $E(B)$ be the complement of $S(B)$ for all $B\in\mathcal{B}$.
Now, using Axiom of choice, let $F(B)$ designate an element of $\mathcal{A}$ containing $B$ for all $B\in\mathcal{B}$.
Next, define $\mathcal{D}\triangleq \{E(B)\cap F(B)\}_{B\in\mathcal{B}}$.
It's clear that $\mathcal{D}$ is a locally finite open refinement of $\mathcal{A}$.
However, i don't really get any motivation for this proof. Why should i take an extra synthetic cover $G$? And how exactly this trick affects the result?
 A: This proof simply isn’t entirely straightforward, and I’m not sure that it’s really possible to make the motivation entirely clear. I’ll try, but I don’t guarantee that you’ll find my explanation satisfying. In the process I’ll modify the proof slightly in a way that I think makes it a little clearer, though for the most part I’ve kept the notation close to yours.
Your hypothesis is that every open cover has a locally finite closed refinement, and you want to show that every open cover has a locally finite open refinement, so you start with an arbitrary open cover $\mathscr{A}$. The only obvious way to apply the hypothesis is to take a locally finite closed refinement $\mathscr{B}$ of $\mathscr{A}$. Somehow you want to use $\mathscr{B}$ to construct a locally finite open refinement of $\mathscr{A}$; presumably this will require you to use the fact that $\mathscr{B}$ is locally finite, so even without seeing how the proof is going to continue it’s natural to write down what this means: for each $x\in X$ there is an open nbhd $G_x$ of $x$ such that $\{B\in\mathscr{B}:G_x\cap B\ne\varnothing\}$ is finite.
However, all of this can be made a little easier to follow if we index the original cover, say $\mathscr{A}=\{A_i:i\in I\}$ for some index set $I$. For each $B\in\mathscr{B}$ choose $i(B)\in I$ such that $B\subseteq A_{i(B)}$, and for each $j\in I$ let $B_j=\bigcup\{B\in\mathscr{B}:i(B)=j\}$; clearly $B_j\subseteq A_j$, and I claim that $B_j$ is closed. This follows from the following useful lemma, whose easy proof I leave to you:

Lemma. If $\mathscr{F}$ is a locally finite collection of closed sets, then $\bigcup\mathscr{F}$ is closed.

Moreover, you can easily check that $\{B_i:i\in I\}$ is still locally finite. Thus, we can replace $\mathscr{B}$ by $\{B_i:i\in I\}$ and still have a locally finite closed refinement of $\mathscr{A}$, so we may as well assume that $\mathscr{B}=\{B_i:i\in I\}$ in the first place. (Note that it is possible that some of the sets $B_i$ are empty.)
The idea now is to expand each $B_i$ a little to an open set $U_i$ in such a way that $B_i\subseteq U_i\subseteq A_i$, and $\{U_i:i\in I\}$ is locally finite. If we can do this, $X\setminus U_i$ will be a closed set disjoint from $B_i$. Moreover, it will have to be a fairly big closed set, since it will have to contain $X\setminus A_i$.
Where can I find big closed sets disjoint from $B_i$? Recall that $\mathscr{B}$ is locally finite so (as noted above) for each $x\in X$ there is an open nbhd $G_x$ of $x$ such that $\{B\in\mathscr{B}:G_x\cap B\ne\varnothing\}$ is finite. This gives me lots of open sets disjoint from $B_i$: all but finitely many of the sets $G_x$ are disjoint from $B_i$. However, I want closed sets disjoint from $B_i$. No problem: $\{G_x:x\in X\}$ is clearly an open cover of $X$, so it has a locally finite closed refinement $\mathscr{C}$. Moreover, we can use the same trick that I used on $\mathscr{B}$ to show that we may assume that $\mathscr{C}=\{C_x:x\in X\}$, where $C_x\subseteq G_x$ for each $x\in X$. 
Now let $i\in I$; $B_i$ intersects only finitely many of the sets $G_x$, so $B_i$ intersects only finitely many of the sets $C_x$, which suggests that the rest of $\mathscr{C}$ must cover a lot of $X$. Let $$\mathscr{C}_i^+=\{C_x\in\mathscr{C}:C_x\cap B_i\ne\varnothing\}$$ and $$\mathscr{C}_i^-=\mathscr{C}\setminus\mathscr{C}_i^+=\{C_x\in\mathscr{C}:C_x\cap B_i=\varnothing\}\;.$$ Let $C_i^+=\bigcup\mathscr{C}_i^+$ and $C_i^-=\bigcup\mathscr{C}_i^-$; by the lemma both of these sets are closed.
In particular, $C_i^-$ is a closed set disjoint from $B_i$. It should be a pretty big one, too, since $\mathscr{C}_i^-$ includes all but finitely many members of the cover $\mathscr{C}$. Let $E_i=X\setminus C_i^-$; clearly $E_i$ is open, and $B_i\subseteq E_i\subseteq C_i^+$. Each $\mathscr{C}_i^+$ is finite, so the closed cover $\{C_i^+:i\in I\}$ is still locally finite, and therefore $\{E_i:i\in I\}$ is locally finite. The family $\{E_i:i\in I\}$ therefore has all of the properties that we wanted for $\{U_i:i\in I\}$ except possibly one: it’s not clear that $E_i\subseteq A_i$ for each $i\in I$.
In fact this might not be true, but that’s not really a problem: we can simply let $U_i=E_i\cap A_i$ for each $i\in I$, and $\mathscr{U}=\{U_i:i\in I\}$ will then be the desired locally finite open refinement of $\mathscr{A}$. The only point that may not be instantly obvious is that $\mathscr{U}$ covers $X$, but this follows immediately from the fact that $\mathscr{B}$ covers $X$, and $B_i\subseteq E_i\cap A_i=U_i$ for each $i\in I$.
