If every finite open covering of a metric space $X$ has a refinement of order $\leqslant n$, is it true that every open covering does too?
We say that a covering has order $n$ if $n$ is the largest natural number $m$ such that there are $m+1$ members of the covering which have a non-empty intersection.
And what about $X$ separable?
Until now, this is what I got:
Hypothesis: every finite open covering of a metric space $X$ has a refinement of order $\leqslant n$
Let $\{A_{\alpha}\}_{\alpha \in \Lambda}$ be an open covering of $X$ and $\tilde{\Lambda}=\Lambda \setminus \{\alpha_1, \alpha_2, \ldots, \alpha_m\}$, for some $m \in \mathbb{N}$ fixed. Then, the collection $$A_{\alpha_1}, A_{\alpha_2}, \ldots, A_{\alpha_m}, \bigcup_{\alpha \in \tilde{\Lambda} } A_{\alpha}$$is a finite open covering of $X$, consequently there is a refinement $$\tilde{A}_{\alpha_1}, \tilde{A}_{\alpha_2}, \ldots, \tilde{A}_{\alpha_m}, \tilde{A}= \bigcup_{\alpha \in \tilde{\Lambda} } \tilde{A}_{\alpha}$$with order $\leqslant n$.
It would be sufficient that this union was disjoint, but I can't demand it.
It seens easier when $X$ is separable, since every covering has a countable subcovering (and every subcovering is a refinement), but I couldn't finish either, even in this case.