TITLE: Anti-compactness
QUESTION [6 upvotes]: Let $(X,\tau)$ be a topological space such that
        $\tau\ne\{\emptyset\ X\}.\ $
We call an open cover $\mathcal{U}$ of $(X,\tau)$ proper if
$\ X\notin \mathcal{U}.\ $ Moreover we say that $(X,\tau)$ is

anti-compact if it does not have a finite proper cover;
anti-paracompact if for every proper cover $\mathcal{U}$ there is $x\in X$ such that every neighborhood intersects infinitely many members of $\mathcal{U}$;
anti-metacompact if for every proper cover $\mathcal{U}$ there is $x\in X$ such that $x$ is a member of infinitely many members of $\mathcal{U}$. 

We have anti-metacompact $\Rightarrow$ anti-paracompact $\Rightarrow$ anti-compact.
Do any of the converse implications hold?

REPLY [3 votes]: Let me provide another proof of @EricWofsey's theorem prompted by Dominic's Question.

Theorem (Eric Wofsay)   Let $\ X\ $ be an anti-compact space,
and let $\mathcal V\ $ be a proper cover. Then every $\ x\in X\ $ belongs to infinitely many members of $\ \mathcal V$.
Proof   Let $\ X\ $ be an anti-compact space,
and let $\mathcal V\ $ be a proper cover, and let $\ x\in X\ $
be such that family $\ \mathcal K:=\{ G\in\mathcal V: x\in G\}\ $
is finite (a proof by contradiction).
Next, let
$\ \mathcal M:=\mathcal V\setminus\mathcal K,\ $ and also
$\ K:=\bigcup\mathcal K\ $ and $\ M:=\bigcup\mathcal M.\ $
Then $\ K\ne X\ne M.\ $ We have a $2$-element open proper covering
$\ \{K\ M\},\ $ which is a contradiction. End of Proof