TITLE: Minimal refinements of open covers of $T_2$-spaces
QUESTION [9 upvotes]: 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.

REPLY [5 votes]: I believe that $\omega_1$ with the order topology is a counter example. Call a subset of $\omega_1$ bounded if it is contained in some initial interval $[0,\alpha]$ and unbounded otherwise. 
Lemma: Any cover of $\omega_1$ by bounded open subsets is not minimal.
I am sure that this is standard, but here is a proof. It contains way too many notations and is a bit clumsy, I am afraid, but I hope it is readable.
We can assume that such a cover is is of cardinality $\omega_1$.
Let thus $\mathcal{V} = \{V_\alpha\,:\,\alpha\in\omega_1\}$ be a cover of $\omega_1$ by bounded open subsets.
By a standard argument the sets below are closed and unbounded in $\omega_1$:
\begin{align*} C_1 &= \{\alpha\,:\, \cup_{\beta<\alpha} V_\beta\subset[0,\alpha)\}\\
C_2 &= \{\alpha\,:\, \cup_{\beta<\alpha} V_\beta\supset[0,\alpha)\}
\end{align*}
Hence their intersection $C = \{\alpha\,:\, \cup_{\beta<\alpha} V_\beta =[0,\alpha)\}$ is closed and unbounded as well. We may assume that each member of $C$ is a limit ordinal. When $\alpha\in \omega_1$, 
let $\alpha^*$ denote the smallest element of $C$ which is $>\alpha$. For each $\alpha\in C$ choose $V_{s(\alpha)}\ni\alpha$ with $s(\alpha)\in[\alpha,\alpha^*)$. Since $V_{s(\alpha)}$ is open and $\alpha$ is limit, $V_{s(\alpha)}$ contains some interval $[\gamma(\alpha),\alpha]$. By Fodor's Lemma, there is some $\gamma\in\omega_1$ such that the set $S$ of $\alpha\in C$ such that $\gamma(\alpha) = \gamma$ is stationary and in particular unbounded. For any $\alpha\in S$ with $\alpha>\gamma^*$, $V_{s(\alpha)}$ contains in particular the interval $[\gamma^*,\alpha]$. 
Fix now $\eta\in[\gamma^*,\gamma^{**})$. Choose $\alpha \ge \gamma^{**}$ in $S$. Then 
$$ V_\eta \subset V_{s(\alpha)} \cup \left(\cup_{\beta<\gamma^*}V_\beta\right).
$$
Notice that $\gamma^*\le\eta<\alpha\le s(\alpha)$, hence the indices in the union at the righthand side do not contain $\eta$. It follows that $\cup (\mathcal{V} \backslash \{V_\eta\}) = \omega_1$.
Corollary: The cover of $\omega_1$ by the intervals $[0,\alpha]$, $\alpha\in\omega_1$, has no refinement that is a minimal cover.