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 $\mathcal U$ be an open cover of $S$.

Let $\mathcal V$ be a point finite open refinement of $\mathcal U$.

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

Define a mapping $H$ on $\mathcal V$ thus:


 * $\forall V \in \mathcal V: H \left({V}\right) = \left\{{U \in \mathcal U: V \subseteq U}\right\}$

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

Call this image $I$.

Since $\mathcal V$ is a refinement of $\mathcal U$, $\varnothing \notin I$.

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

Then $G = c \circ H: \mathcal V \to \mathcal U$ is a mapping such that:
 * $\forall V \in \mathcal V: V \subseteq G \left({V}\right)$

Then $\mathcal Q = G \left({\mathcal V}\right)$ is countable by Image of Countable Set under Mapping is Countable.

Each element of $\mathcal Q$ is an element of $\mathcal U$ by the definition of $G$.

Let $x \in S$.

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

Then $x \in V \subseteq G \left({V}\right) \in \mathcal Q$.

Thus $\mathcal Q$ is a countable subcover of $\mathcal U$.

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