Separable Metacompact Space is Lindelöf/Proof 2

Proof
$T$ is metacompact every open cover of $S$ has an open refinement which is point finite.

$T$ is a Lindelöf space if every open cover of $S$ has a countable subcover.

Having established the definitions, we proceed.

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

Let $\VV$ be a point finite open refinement of $\UU$.

By Point Finite Set of Open Sets in Separable Space is Countable, $\VV$ is countable.

Define a mapping $H$ on $\VV$ thus:


 * $\forall V \in \VV: \map H V = \set {U \in \UU: V \subseteq U}$

By Image of Countable Set under Mapping is Countable, the image of $H$ is countable.

Call this image $I$.

Since $\VV$ is a refinement of $\UU$, $\O \notin I$.

By the Axiom of Countable Choice, $I$ has a choice function $c$.

Then $G = c \circ H: \VV \to \UU$ is a mapping such that:
 * $\forall V \in \VV: V \subseteq \map G V$

Then $\QQ = \map G \VV$ is countable by Image of Countable Set under Mapping is Countable.

Each element of $\QQ$ is an element of $\UU$ by the definition of $G$.

Let $x \in S$.

Then since $\VV$ is a cover for $S$:
 * $\exists V \in \VV: x \in V$

Then $x \in V \subseteq \map G V \in \QQ$.

Thus $\QQ$ is a countable subcover of $\UU$.

Thus each open cover of $S$ has a countable subcover, so $T$ is a Lindelöf space.