Open Cover with Closed Locally Finite Refinement is Even Cover/Lemma 1

Theorem
Let $T = \struct{X, \tau}$ be a topological Space.

Let $\UU$ be an open cover of $T$.

Let $\AA$ be a closed locally finite refinement of $\UU$.

For each $A \in \AA$, let $U_A \in \UU$ such that $A \subseteq U_A$.

For each $A \in \AA$, let:
 * $V_A = \paren{U_A \times U_A} \cup \paren{\paren{X \setminus A} \times \paren{X \setminus A}}$

For each $x \in X, A \in \AA$, let:
 * $V_A \sqbrk x =\set{ y∈X : \tuple{x,y} ∈ V_A}$

Let:
 * $V = \ds \bigcap_{A \in \AA} V_A$

For each $x \in X$, let:
 * $V \sqbrk x =\set{ y∈X : \tuple{x,y} ∈ V}$

Then:
 * $\set{V \sqbrk x : x \in X}$ is a refinement of $\UU$

Proof
Let $x \in X$.

By definition of refinement:
 * $\AA$ is a cover of $X$

By definition of cover:
 * $\exists A \in \AA : x \in A$

Lemma 2
From lemma 2
 * $V \sqbrk x \subseteq V_A \sqbrk x = U_A \in \UU$

Since $x$ was arbitrary, then:
 * $\forall x \in X : \exists U \in \UU : V \sqbrk x \subseteq U$

Also, we have:
 * $\forall x \in X : x \in V \sqbrk x$

It follows that $\set{V \sqbrk x : x \in X}$ is a refinement of $\UU$ by definition.