Fully T4 Space is T4 Space

Theorem
Let $T = \left({X, \tau}\right)$ be a fully $T_4$ space.

Then $T$ is a $T_4$ space.

Proof
From the definition, $T$ is fully $T_4$ iff every open cover of $X$ has a star refinement.

Let $\mathcal U$ be an open cover for $T$.

Then from the definition, there exists a be a cover $\mathcal V$ for $T$ such that:
 * $\displaystyle \forall x \in S: \exists U \in \mathcal U: \left({\bigcup \left\{{V \in \mathcal V: x \in V}\right\} }\right) \subseteq U$

$\left({X, \tau}\right)$ is a $T_4$ space iff:


 * $\forall A, B \in \complement \left({\tau}\right), A \cap B = \varnothing: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \varnothing$

That is, for any two disjoint closed sets $A, B \subseteq X$ there exist disjoint open sets $U, V \in \tau$ containing $A$ and $B$ respectively.

That is:
 * $\left({X, \tau}\right)$ is $T_4$ when any two disjoint closed subsets of $X$ are separated by neighborhoods.