This is a follow-up question to an older question.

Let $X\neq \emptyset$ be a set. We say that ${\cal C}\subseteq {\cal P}(X)$ is a cover if $\bigcup {\cal C} = X$, and we call ${\cal C}$ linear if $|A \cap B| \leq 1$ whenever $A\neq B \in {\cal C}$.

We say ${\cal C}$ is minimal if for all $A \in {\cal C}$ the set ${\cal C}\setminus \{A\}$ is no longer a cover.

If $X$ is a non-empty set and ${\cal C} \subseteq {\cal P}(X)$ is a linear cover, is there necessarily a minimal cover ${\cal C}'\subseteq {\cal C}$?

This is a follow-up question to an older question.

Let $X\neq \emptyset$ be a set. We say that ${\cal C}\subseteq {\cal P}(X)$ is a cover if $\bigcup {\cal C} = X$, and we call ${\cal C}$ linear if $|A \cap B| \leq 1$ whenever $A\neq B \in {\cal C}$.

We say ${\cal C}$ is minimal if for all $A \in {\cal C}$ the set ${\cal C}\setminus \{A\}$ is no longer a cover. Note that ${\cal C}$ is minimal if and only if for every $A \in {\cal C}$ there is $x \in A$ such that $A$ is the only member of ${\cal C}$ containing $x$.

If $X$ is a non-empty set and ${\cal C} \subseteq {\cal P}(X)$ is a linear cover, is there necessarily a minimal cover ${\cal C}'\subseteq {\cal C}$?