Separable Metacompact Space is Lindelöf/Proof 1

Proof
$T$ is separable there exists a countable subset of $S$ which is everywhere dense.

$T$ 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 $T = \left({S, \tau}\right)$ be separable and metacompact.

there exists an open cover $\UU$ of $S$ which has no countable subcover.

As $T$ is metacompact, $\UU$ has an open refinement $\VV$ which is point finite.

By nature of $\UU$, which has no countable subcover, $\VV$ is uncountable.

By hypothesis, $T$ is separable.

Let $\SS$ be a countable subset of $S$ which is everywhere dense.

Then each $V \in \VV$ contains some $s \in \SS$.

So some $s \in \SS$ is contained in an uncountable number of elements of $\VV$.

Thus, by definition, $\VV$ is not point finite.

Thus no uncountable open refinement $\VV$ of $\UU$ exists which is point finite.

It follows that $\VV$ must be countable.

Thus $\UU$ has a countable subcover.

That is, by definition, $T$ is a Lindelöf space.