Separable Metacompact Space is Lindelöf/Proof 1

Theorem
Let $T = \left({X, \tau}\right)$ be a separable topological space which is also metacompact.

Then $T$ is a Lindelöf space.

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

$T$ is metacompact iff every open cover of $X$ has an open refinement which is point finite.

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

Having established the definitions, we proceed.

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

Suppose, in order to derive a contradiction there exists an open cover $\mathcal U$ of $X$ which has no countable subcover.

As $T$ is metacompact, $\mathcal U$ has an open refinement $\mathcal V$ which is point finite.

By nature of $\mathcal U$, which has no countable subcover, $\mathcal V$ is uncountable.

By hypothesis, $T$ is separable.

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

Then each $V \in \mathcal V$ contains some $s \in \mathcal S$.

So some $s \in \mathcal S$ is contained in an uncountable number of elements of $\mathcal V$.

Thus, by definition, $\mathcal V$ is not point finite.

Thus no uncountable open refinement $\mathcal V$ of $\mathcal U$ exists which is point finite.

It follows that $\mathcal V$ must be countable.

Thus $\mathcal U$ has a countable subcover.

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