Queries on munkres lemma 41.3 In Munkres Topology chapter 6 lemma 41.3(page 254)
Lemma 41.3:
Let $X$ be regular. Then the following conditions on $X$ are equivalent:
Every open covering of $X$ has a refinement that is:

*

*An open covering of $X$ and countably locally finite

*A covering of $X$ and locally finite

*A closed covering of $X$ and locally finite

*An open covering of $X$ and locally finite

In the proof of (2)$\Rightarrow$(3) (and (3)$\Rightarrow$(4)), Munkres constructs a covering $\mathscr{C}$ and seems to assume that it is countably locally finite as it uses (2) (and (3))

(2)$\Rightarrow$(3):
Let $\mathscr{A}$ be an open covering of $X$. Let $\mathscr{B}$ be the collection of all open sets $U$ of $X$ such that $\overline{U}$ is contained in an element of $A$. By regularity, $B$ covers $X$. Using (2), we can find a refinement $\mathscr{C}$ of $\mathscr{B}$ that covers $X$ and is locally finite.

In order to use (2) it seems like $\mathscr{B}$ is assumed to be countably locally finite. But suppose we let $\mathscr{B}$ be the collections of open sets in the lower limit topology that is contained in $[0,1)$, it doesn't seem like this collection is countably locally finite.
Similarly,

(3)$\Rightarrow$(4):
For each point $x$ of $X$, there is a neighborhood of $x$ that intersects only finitely many elements of $\mathscr{B}$. The collection of all open sets that intersect only finitely many elements of $\mathscr{B}$ is thus an open covering of $X$. Using (3) again, let $\mathscr{C}$ be a closed refinement of this covering that covers $X$ and is locally finite.

Which also assumes $\mathscr{B}$ is countably locally finite
My doubt is how is $\mathscr{B}$ is countably locally finite in both cases, since it doesn't seem to be in the specific case mentioned above.
Edit 1: The structure of the proof is pretty clear, it's just that using (2) or (3) assumes (1), which states that the covering have a refinement that is countably locally finite, not all coverings in regular spaces are countably locally finite, for instance $R_\ell^2$ is regular but not paracompact
Edit 2: So the assumption at first was that every open covering, not just a single open covering, so $\mathscr{B}$ by assumption is countably locally finite
 A: The structure of the proof is a classical "round trip" of implications: so we assume $(2)$ holds ("every open cover has a locally finite refinement"; where this refinement is just any cover of $X$, the elements can be any type of set) and show $(3)$: "every open cover has a locally finite closed refinement". 
So to show $(3)$, we start with an open cover $\mathcal{A}$ (and we can assume $(2)$ as a fact), but first we define an other related cover (not locally finite, in general ) $$\mathcal{B}=\{U: \text{ open }: \exists A \in \mathcal{A}: \overline{U}\subseteq A\}$$
This is a cover indeed: let $x\in  X$, as $\mathcal{A}$ is an open cover there is some $A_x \in \mathcal{A} $ with $x \in A_x$. By regularity there is an open subset $U$ of $X$ such that $x \in U \subseteq \overline{U} \subseteq A_x$. Then $U \in \mathcal{B}$ as witnessed by $A_x$ and $x \in U$.
Now apply $(2)$ to this cover $\mathcal{B}$, and we get a locally finite $\mathcal{C}$ that is a cover of $X$ and refines $\mathcal{B}$. Then he probably goes on to show that
$$\mathcal{C}'=\{\overline{C}: C \in \mathcal{C}\}$$
is a closed locally finite cover of $X$, where we can maybe rely on some old fact that when a family of set is locally finite, so is the set of its closures.
That it is indeed a cover and refines the original $\mathcal{A}$ is easy to check from definitions.
$(3)$ to $(4)$ is similar: we start with an open cover, define a related open cover, and apply $(3)$ to that related cover, and "manipulate" the (locally finite) result to show $(4)$, and get a refinement of the right type for the original open cover.
A: For any conditional statement, $P \Rightarrow Q$, the statement evaluates to true for all possible truth values of $P,Q$ except $T \Rightarrow F = F$, but $F \Rightarrow Q = T, T\Rightarrow T = T$. So, the assumption is warranted to prove the conditional statement. You need only consider the case when $P$ is true and prove the implication is that $Q$ is also true.
In your example, if you want to prove that (2) implies (3), then assume (2) is true, and use its truth to prove that (3) is also true.
