Paracompact and Normal spaces Lemma: Let $X$ be a paracompact space. Let $A,B$ two closed subset of $X$ such that $A \cap B=\emptyset$. If $\forall x \in B$ there exists disjoint open sets $U_x$ and $V_x$ such that $A\subseteq U_x$ and $x \in V_x$, then there exists disjoint open set $U$ and $V$ such that $A\subseteq U$ and $B \subseteq V$.
Proof:
The open cover $\{X\setminus B\}\cup\{V_x: x \in B\}$ of $X$ has locally finite open refinement $\{W_\gamma\mid \gamma \in \Gamma\}$.
Let $Γ_1=\{\gamma \in \Gamma: W_\gamma \subseteq V_x  \,\text{for some}\, x\in B\}$.
Since $x \notin \bigcup_{\gamma \in Γ_1} W_\gamma \implies x\notin w_γ,∋γ∈Γ_1⟶ x∉v_x,∀x∈B⟶ x∉B$, then $B⊆⋃_{γ∈Γ_1} w_γ$.

I am trying to understand the first part of this proof. I am stuck at this part $x∉w_γ,∋γ∈Γ_1⟶ x∉v_x,∀x∈B$ because $w_γ⊆v_x$ so why is $x∉w_γ,∋γ∈Γ_1⟶ x∉v_x,∀x∈B$ true?
 A: The author wants to show that $$B \subseteq \bigcup_{\gamma \in \Gamma_1} W_\gamma\tag{1}$$
So take $x \in B$. This is covered by some $W_\gamma$ (so $x \in W_\gamma$ for some $\gamma \in \Gamma$) (from the fact that the $W_\gamma$ cover $X$). This $W_\gamma$ is not a subset of $X\setminus B$ (as witnessed by $x$ itself) and as it is a refinement of the unnamed used cover, we must have that it is a subset of $V_p$ for some $p \in B$ by definition. This then shows that by definition in fact $\gamma \in \Gamma_1$ and so the inclusion $(1)$ has been shown. We use nothing but the definitions and common sense.
IMO all these strange formulaic notation that you or your text is using just obscures these straightforward facts. Just reason logically and more importantly, write natural language.
As to the text argument write-up:
There are two types of $\gamma$: those such that $W_\gamma$ is contained in some $V_x, x \in B$ or those whose $W_\gamma$ is contained in $X\setminus B$; $\Gamma_1$ collects those of the first type. So if $x \notin \bigcup_{\gamma \in \Gamma_1} W_\gamma$, it must be covered by some $W_\gamma$ (they form a cover of $X$) and by assumption is not of the first type. Hence $\gamma \notin \Gamma_1$. So $x \in W_\gamma \subseteq X\setminus B$ and so $x \notin B$. The proof's intermediate claim that $x$ is in none of the $V_y, y \in B$ is nonsense, it could very well be in one (say if it's outside $B$ but "close" to it). The claim is unnecessary for the final inclusion which we do need.
