Characterization of Paracompactness in T3 Space/Statement 2 implies Statement 3

Theorem
Let $T = \struct{X, \tau}$ be a $T_3$ Space.

If every open cover of $T$ has a locally finite refinement then:
 * every open cover of $T$ has a closed locally finite refinement

Proof
Let every open cover of $T$ have a locally finite refinement.

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

Let $\VV = \set{V \in \tau : \exists U \in \UU : V^- \subseteq U}$ where $V^-$ denoted the closure of $V$ in $T$.

$\VV$ is an Open Cover of $T$
Let $x \in S$.

By definition of open cover:
 * $\exists U \in \UU : x \in U$

From Characterization of T3 Space:
 * $\exists V \in \tau : x \in V : V^- \subseteq U$

Hence:
 * $V \in \VV$

By definition, $\VV$ is an open cover.

By assumption:
 * there exists a locally finite refinement $\AA$ of $\VV$.

Let:
 * $\BB = \set{A^- : A \in \AA}$

From User:Leigh.Samphier/Topology/Closures of Elements of Locally Finite Set is Locally Finite:
 * $\BB$ is locally finite

$\BB$ is an Open Cover of $T$
Let $x \in S$.

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

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

From Set is Subset of its Topological Closure:
 * $A \subseteq A^-$

By definition of subset:
 * $x \in A^-$

By definition of $\BB$:
 * $A^- \in \BB$

By definition, $\BB$ is an open cover.

$\BB$ is a Refinement of $\UU$
Let $B \in \BB$.

By definition of $\BB$:
 * $\exists A \in \AA : A^- = B$

By definition of refinement:
 * $\exists V \in \VV : A \subseteq V$

From Set Closure Preserves Set Inclusion:
 * $B = A^- \subseteq V^-$

By definition of $\VV$:
 * $\exists U \in \UU : V^- \subseteq U$

From Subset Relation is Transitive:
 * $B \subseteq U$

It follows that $\BB$ is a refinement of $\UU$.

Since $\UU$ was an arbitrary open cover of $T$ it follows that:
 * every open cover of $T$ has a closed locally finite refinement.