Open cover not containing a certain subcover

Is there an infinite topological space $(X,\tau)$ with the following property?

There is an open cover ${\cal U}^*$ such that

• $X\notin {\cal U}^*$;
• every finite subset $F\subseteq X$ is contained in some member of ${\cal U}^*$ and
• for every subcover ${\cal V} \subseteq {\cal U}^*$ there is $x\in X$ such that $\{V\in {\cal V}: x\notin V\}$ is infinite.
• How about an uncountable discrete space, with the open cover consisting of all finite subsets? – Jeremy Rickard Oct 1 '15 at 9:16
• @JeremyRickard: Actually, I think your example works for countably infinite spaces as well. – Jason Starr Oct 1 '15 at 9:38
• @JasonStarr For a discrete countably infinite set you could take a subcover consisting of a nested sequence $U_0\subset U_1\subset\dots$ of finite sets. – Jeremy Rickard Oct 1 '15 at 9:45
• You are right. Countably infinite spaces do not work. – Jason Starr Oct 1 '15 at 9:55

As a counterexample, consider any topological space $X$ which is not metacompact.
We recall that a topological space $X$ is metacompact each each open cover of $X$ has a point-finite refinement. By Theorem 3.5 in the survey "Covering properties" by D.Burke in the Handbook of Set-Theoretic Topology (1984), a topological space $X$ is metacompact if and only if any directed open cover of $X$ has a point-finite refinement. A cover $\mathcal U$ of $X$ is directed if for any sets $U,V\in\mathcal U$ there is a set $W\in\mathcal U$ containing the union $U\cup V$.
By this characterization, each topological space $X$ which is not metacompact admits a directed open cover $\mathcal U$ without point-finite refinement. Since $\mathcal U$ is dierected, each finite subset $F\subset X$ is contained in some set $U\in\mathcal U$. Since $\mathcal U$ has no point-finite refinement, any subcover $\mathcal V$ of $\mathcal U$ is not point-finite, which means that the set $\{V\in\mathcal V:x\in V\}$ is infinite for some $x\in X$.
For an example of a space $X$ which is not metacompact, see Example 4.4 of the survey of Burke. According to this example any Mrowka space $\psi(\mathbb N)$ is not meta-Lindelof and hence not metacompact.