Anti-compactness

A space is anti-compact iff it has no proper covers consisting of two sets, or equivalently if the intersection of any two nonempty closed sets is nonempty. This is equivalent to the specialization order being directed downwards.

We can use this to prove any anti-compact space is anti-metacompact, so your three conditions are equivalent. Suppose $X$ is anti-compact, and let $\mathcal{U}$ be a proper cover. Choose some $U_0\in\mathcal{U}$ and some $x_0\in U_0$. Let $y_0\in X\setminus U_0$, let $x_1$ be a common lower bound of $x_0$ and $y_0$ with respect to the specialization order, and choose $U_1\in \mathcal{U}$ containing $x_1$. Let $y_1\in X\setminus (U_0\cup U_1)$, $x_2$ be a common lower bound of $x_1$ and $y_1$, and choose $U_2\in \mathcal{U}$ containing $x_2$. Continuing by induction, we get a decreasing sequence $x_0>x_1>x_2>\dots$ and distinct sets $U_n\in \mathcal{U}$ such that $x_n\in U_n$ for all $n$. It follows that $x_0$ is in every $U_n$ and so $X$ is anti-metacompact. In fact, since $x_0$ was arbitrary, every element of $X$ is in infinitely many members of $\mathcal{U}$.

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