Let $(X,\tau)$ be a topological space, and let ${\cal U}$ be an open cover. We say that ${\cal U}$ is *thick* if for all $x\in X$ we have $$|\{V\in {\cal U}: x\in V\}| = |X|.$$
Is there a Hausdorff space $(X,\tau)$ with $|X|>1$ and an open cover ${\cal U}$ of $X$ such that every [refinement](https://en.wikipedia.org/wiki/Cover_(topology)#Refinement) of ${\cal U}$ is thick?



Thick refinements of covers
This is impossible for any $T\_1$ space $X$. For suppose $X$ is $T\_1$ and $\mathcal U$ is an open cover of $X$, and fix $x \in X$. If $U$ is some member of $\mathcal U$ containing $x$, then $\{U\} \cup \{V \setminus \{x\} : V \in \mathcal U,\, V \neq U\}$ is a refinement of $\mathcal U$, but it is not thick because only one set in this refinement contains $x$.


