Paracompact Space is Metacompact

Theorem
Let $T = \left({X, \vartheta}\right)$ be a paracompact space.

Then $T$ is metacompact.

Proof
From the definition, $T$ is paracompact iff every open cover of $X$ has an open refinement which is locally finite.

Consider some open cover $\mathcal U$ of $X$.

Let $x \in X$.

Then there exists some neighborhood $\exists N_x$ of $x$ which intersects only finitely many elements of $\mathcal U$.

Thus $x$ itself can be in only finitely many elements of $\mathcal U$.

Hence $T$ must be, by definition, metacompact.