Existence of minimal subcover for any open cover of a metric space 
Suppose $(X,d)$ is a metric space. Does every open cover of $X$ have a minimal subcover with respect to inclusion?

In other words:

If $\mathcal{O}$ is an open cover of a metric space $X$, then does there exist an open cover $\mathcal{U} \subseteq \mathcal{O}$ such that, if $\mathcal{U}' \subsetneq \mathcal{U}$, then $\mathcal{U}'$ does not cover $X$ ?

 A: As other answers have pointed out, there are easy counterexamples. What is true is that if $\mathscr{U}$ is an open cover of a metric space $X$, then $\mathscr{U}$ has an irreducible open refinement: that is, there is an open cover $\mathscr{R}$ of $X$ such that 


*

*for each $R\in\mathscr{R}$ there is a $U\in\mathscr{U}$ such that $R\subseteq U$, and  

*for each $R\in\mathscr{R}$, the family $\mathscr{R}\setminus\{R\}$ no longer covers $X$.


This is a consequence of two well-known theorems. First, every metric space is paracompact, so every open cover of a metric space has a locally finite open refinement. Secondly, every point finite cover of a set (and a fortiori every locally finite cover) has an irreducible subcover. 
A: Take any $x\in X$, and consider the covering by the sets $\{y\in X:d(x,y)<n\}$ for all positive integers $n$. If $X$ is unbounded (with respect to the metric $d$), then this open cover has no minimal subcover.
A: Let $X=\mathbb R$ and $U_n=(-n,n)$. Then $\{U_n\}$ covers $X$ but it doesn't have a minimal sub-cover.
