This is not even a partial answer, just an oversized comment. I consider only the special case where the set $X$ is countable and the intersecting cover $\mathcal C$ consists of finite sets. I am not able to settle even this special case, but I would like to record some equivalences which might be useful in settling it.
Let $\mathbf F$ be the set of all intersecting covers of $\mathbb N$ consisting of finite sets.
Lemma. For any $\mathcal C\in\mathbf F$ there exists $\mathcal C_1\in\mathbf F$ such that $\mathcal C_1\preceq\mathcal C$ and $\mathcal C_1$ is an antichain (no element of $\mathcal C_1$ is a subset of another).
Proof. Let $\mathcal M$ be the set of all minimal elements of $\mathcal C$ and let $\mathcal C_0=\{A\cup\{x\}:A\in\mathcal M,\ \{A\cup\{x\}\}\preceq C\}$. Then $\mathcal M\subseteq\mathcal C_0\preceq\mathcal C$ and $\mathcal C_0\in\mathbf F$. Let $\mathcal C_1$ be the set of all maximal elements of $\mathcal C_0$. Plainly $\mathcal C_1$ is an antichain and an intersecting family, and $\mathcal C_1\preceq\mathcal C$; I have to show that $\mathcal C_1$ covers $\mathbb N$. In fact I will show that there is no infinite chain in $\mathcal C_0$, whence every element of $\mathcal C_0$ is contained in a maximal element.
Assume for a contradiction that $\mathcal C_0$ contains an infinite chain
$$A_1\cup\{x_1\}\subsetneqq A_2\cup\{x_2\}\subsetneqq A_3\cup\{x_3\}\subsetneqq\cdots$$
where $A_n\in\mathcal M$. Now $x_n\notin A_n$ for $n\ge2$, since $A_n\cup\{x_n\}\notin\mathcal M$ for $n\ge2$. Moreover, since $A_1\subseteq A_n\cup\{x_n\}$, while $A_1\not\subseteq A_n$ for $n\ge3$, we have $x_n\in A_1$ for $n\ge3$. Since $A_1$ is finite, by the pigeonhole principle we have $x_m=x_n=x$ for some $m,n$ with $3\le m\lt n$. But now, since $A_m\cup\{x\}\subsetneqq A_n\cup\{x\}$ and $x\notin A_m$, we have $A_m\subsetneqq A_n$, which is absurd.
Theorem. The following statements are equivalent:
(1) For any $\mathcal C\in\mathbf F$ there exists $\mathcal C'\in\mathbf F$, with $\mathcal C'\preceq\mathcal C$, such that $$\mathcal C'\succeq\mathcal B\in\mathbf F\implies\mathcal C'\preceq\mathcal B;$$
i.e., $\mathcal C'$ is refinement-minimal.
(2) For any $\mathcal C\in\mathbf F$ there exists $\mathcal C'\in\mathbf F$, with $\mathcal C'\preceq\mathcal C$, such that $$\mathcal C'\succeq\mathcal B\in\mathbf F\implies\mathcal C'\subseteq\mathcal B.$$
(3) For any $\mathcal C\in\mathbf F$ and any $x\in\mathbb N$ there exist $\mathcal C'$ and $A$, with $x\in A\in\mathcal C'\in\mathbf F$ and $\mathcal C'\preceq\mathcal C$, such that
$$\mathcal C'\succeq\mathcal B\in\mathbf F\implies A\in\mathcal B.$$
Proof. Plainly (2)$\implies$(1) & (3); I will show that (1)$\implies$(2) and (3)$\implies$(2).
Proof of (1)$\implies$(2). Suppose $\mathcal C\in\mathbf F$. By (1) there exists $\mathcal C_0\in\mathbf F$ with $\mathcal C_0\preceq\mathcal C$ such that $\mathcal C_0$ is refinement-minimal. By the lemma there exists $\mathcal C'\in\mathbf F$ such that $\mathcal C'\preceq\mathcal C_0$ and $\mathcal C'$ is an antichain. Now, if $\mathcal C'\succeq\mathcal B\in\mathbf F$, then $\mathcal C'\preceq\mathcal B$; since $\mathcal C'$ is an antichain, it follows that $\mathcal C'\subseteq\mathcal B$.
Proof of (3)$\implies$(2). Let $\mathcal C_0=\mathcal C$. Using (3) we can construct $\mathcal C_n\in\mathbf F$ and $A_n\in\mathcal C_n$ for $n\in\mathbb N$ so that $\mathcal C_n\preceq\mathcal C_{n-1}$ and $n\in A_n$, and so that $\mathcal C_n\succeq\mathcal B\in\mathbf F\implies A_n\in\mathcal B$. It follows that $A_n\in\mathcal A_m$ for all $m\ge n$. Let $\mathcal C'=\{A_n:n\in\mathbb N\}$. Then $\mathcal C'\in\mathbf F$, and $\mathcal C'\preceq\mathcal C_n$ for all $n$, and $\mathcal C'\succeq\mathcal B\in\mathbf F\implies\mathcal C'\subseteq\mathcal B$.
Remark. I wanted to include the following statement but I couldn't prove that it's equivalent to the others:
(4) For any $\mathcal C\in\mathbf F$ and any $x\in\mathbb N$ there exist $\mathcal C'$ and $A$, with $x\in A\in\mathcal C'\in\mathbf F$ and $\mathcal C'\preceq\mathcal C$, such that
$$\mathcal C'\succeq\mathcal B\in\mathbf F\implies\{A\}\preceq\mathcal B.$$
