Let us recall that $\mathfrak u$ is the smallest cardinality of a base of a free ultrafilter on $\omega$. It is known (and easy to see) that $\omega_1\le\mathfrak u\le\mathfrak c$.
Example. There exists an intersecting cover $\mathcal C$ of a set $X$ of cardinality $|X|=\mathfrak u$ such that every intersecting cover $\mathcal C_0$ of $X$ with $\mathcal C_0\preceq \mathcal C$ is not refinement-minimal.
Proof. Take any free ultrafilter $\mathcal U$ on $\omega$ with base $\mathcal B$ of cardinality $|\mathcal B|=\mathfrak u$. Consider the set $X=\omega\cup\mathcal B$ and observe that $\mathcal C:=\{\{B\}\cup B:B\in\mathcal B\}$ is an intersecting cover of $X$.
Now take any intersecting cover $\mathcal C_0$ of $X$ with $\mathcal C_0\preceq \mathcal C$.
Claim 1. For every set $C\in\mathcal C_0$, the set $C\cap \omega$ belongs to the ultrafilter $\mathcal U$.
Proof. In the opposite case, the set $\omega\setminus C$ belongs to the ultrafilter $\mathcal U$. Since $\mathcal C_0\preceq\mathcal C$, there exists a set $B\in\mathcal B$ such that $C\subseteq \{B\}\cup B$. Since $\omega\setminus C\in\mathcal U$ and $B\in\mathcal B\subseteq\mathcal U$, there exists a set $B'\in\mathcal B\setminus\{B\}$ such that $B'\subseteq B\setminus C$. Since $\mathcal C_0$ is a cover of $X=\mathcal B\cup\omega$, there exists an element $C'\in\mathcal C_0$ such that $B'\in C'$. Since $\mathcal C_0\preceq \mathcal C$, there exists a set $B''\in\mathcal B$ such that $B'\in C'\subseteq \{B''\}\cup B''$, which implies that $B''=B'$ and hence $C'\subseteq \{B'\}\cup B'$ and $$C\cap C'\subseteq C\cap (\{B\}\cup B)\cap (\{B'\}\cup B')=C\cap B\cap B'\subseteq C\cap (B\setminus C)=\emptyset,$$ which contradicts the intersection property of the family $\mathcal C_0$. $\quad\square$
Now take any set $B\in\mathcal B$ and consider the subfamily $\mathcal C_B:=\{C\in\mathcal C_0:B\in C\}$.
Claim 2. For every $C\in\mathcal C_B$ we have $B\in C\subseteq \{B\}\cup B$.
Proof. Since $\mathcal C_0\preceq \mathcal B$, there exists a set $B'\in\mathcal B$ such that $B\in C\subseteq\{B'\}\cup B'$, which implies $B=B'$ and hence $B\in C\subseteq \{B\}\cup B$. $\quad\square$
Choose any set $C'\in\mathcal C_B$ and consider the set $U':=\omega\cap\bigcup\mathcal C_B$, which belongs to the ultrafilter $\mathcal U$, by Claim 1. Choose a base $\mathcal D$ of the ultrafilter $\mathcal U':=\{U'\cap U:U\in\mathcal U\}$ such that $U'=\bigcup\mathcal D$, $\mathcal D\preceq\mathcal C_B$, but $C'\setminus\{B\}\not\subseteq D$ for every $D\in\mathcal D$. Consider the family $$\mathcal C'_0:=(\mathcal C_0\setminus\mathcal C_B)\cup\{\{B\}\cup D:D\in\mathcal D\}.$$ It is easy to see that $\mathcal C_0'$ is an intersecting cover of $X$ such that $\mathcal C_0'\preceq\mathcal C_0$. Assuming that $\mathcal C_0\preceq\mathcal C_0'$, we can find a set $D\in\mathcal D$ such that $C'\subseteq\{D\}\cup D$ and hence $C'\setminus\{B\}\subseteq D$, which contradicts the choice of the base $\mathcal D$. This contradiction shows that the intersecting cover $\mathcal C_0$ is not refinement-minimal. $\quad\square$
The above Example motivates the following problem (of exchange the uncountable cardinal $\mathfrak u$ by the cardinal $\omega$):
Problem. Is there an intersecting cover without refinement-minimal refinements on a countable set?