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.