Let $(X,\tau)$ be a topological space. We say ${\cal U}\subseteq \tau$ is an open cover if

• $\bigcup {\cal U} = X$, and
• $X\notin {\cal U}$.

${\cal U}$ is minimal if for all $U_0\in {\cal U}$ we have $\bigcup \big({\cal U}\setminus \{U_0\}\big) \neq X$. Clearly, every $T_1$-space on more than $1$ point possesses a minimal cover: pick $x\neq y\in X$ and let ${\cal U} = \big\{X\setminus\{x\}, X\setminus\{y\}\big\}$.

Question. Given any open cover of a Hausdorff space $(X,\tau)$ with $|X|>1$, does it have a refinement that is a minimal cover?

Note. I did not ask for subcovers in the question because of the following example: Let $X = \mathbb{R}$ with the Euclidean topology, then ${\cal U} = \big\{ \{x\in\mathbb{R}: x < n\} : n\in\mathbb{N}, n\geq 1 \big\}$ does not have a minimal subcover.