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 space $(X,\tau)$ and an open cover ${\cal U}$ such that every refinement of ${\cal U}$ is thick?

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 of ${\cal U}$ is thick?