Partial answer. I show that a minimal subcover exists in two special cases, namely, if $\mathcal C$ consists of $2$-element sets, or if $X$ is countable.
Theorem 1. A cover consisting of $2$-element sets has a minimal subcover.
Proof. In other words, I claim that a graph $G$ with no isolated vertices has a spanning subgraph $H$ which is a star-forest with no isolated vertices, i.e., each component of $H$ is a tree of diameter $1$ or $2$. First, choose a maximal collection of disjoint edges. Next, for each vertex $x$ which is covered by none of the chosen edges, add one edge incident with $x$. Now we have a spanning subgraph in which each component is a tree of diameter $1$, $2$, or $3$. Finally, delete the central edge from each component of diameter $3$.
Definition. If $\mathcal C\subseteq\mathcal P(X)$ and $x\in X$, let $\mathcal C(x)=\{A\in\mathcal C:x\in A\}$.
Lemma. If $\mathcal C$ is a linear cover of $X$, then $$\forall x\in X,\ \exists A\in\mathcal C(x),\ \exists y\in A,\ \forall z\in X\setminus A,\ [\mathcal C(z)\not\subseteq\mathcal C(y)].$$
Proof. Let $\mathcal C$ be a linear cover of $X$, and assume for a contradiction that there is an element $x_0\in X$ such that $$\forall A\in\mathcal C(x_0),\ \forall y\in A,\ \exists z\in X\setminus A,\ [\mathcal C(z)\subseteq\mathcal C(y)].$$ Choose $A_0\in\mathcal C(x_0)$ and then choose $x_1\in X\setminus A_0$ so that $\mathcal C(x_1)\subseteq\mathcal C(x_0)$. Choose $A_1\in\mathcal C(x_1)$. Since $A_1\in\mathcal C(x_0)$ and $x_1\in A_1$, we can choose $x_2\in X\setminus A_1$ so that $\mathcal C(x_2)\subseteq\mathcal C(x_1)$. Choose $A_2\in\mathcal C(x_2)$.
Now $A_1,A_2\in\mathcal C$, and $A_1\ne A_2$ since $x_2\in A_2\setminus A_1$. Moreover $x_0,x_1\in A_1\cap A_2$, and $x_0\ne x_1$ since $x_0\in A_0$ while $x_1\notin A_0$, so $|A_1\cap A_2|\ge2$, contradicting the assumed linearity of $\mathcal C$.
Theorem 2. Every linear cover of $\mathbb N$ has a minimal subcover.
Proof. Let $\mathcal C$ be a linear cover of $\mathbb N$. I aim to construct $A_i\in\mathcal C$ and $x_i\in\mathbb N$ ($i\in\mathbb N$) satisfying the following conditions for all $i,j\in\mathbb N$:
(i) $i\in A_i$;
(ii) $x_i\in A_j\iff A_i=A_j$;
(iii) $\forall z\in\mathbb N\setminus(A_1\cup\cdots\cup A_i),\ [\mathcal C(z)\not\subseteq\mathcal C(x_1)\cup\cdots\cup\mathcal C(x_i)]$.
It will follow from (i) and (ii) that $\{A_i:i\in\mathbb N\}$ is a minimal subcover of $\mathcal C$.
Let $n\ge0$ and suppose that $A_1,\dots,A_n\in\mathcal C$ and $x_1,\dots,x_n\in\mathbb N$ have been defined so that (i)–(iii) are satisfied for $i,j\le n$. We consider two cases.
The trivial case: $n+1\in A_1\cup\cdots\cup A_n$.
Choose $k\le n$ so that $n+1\in A_k$; let $A_{n+1}=A_k$ and $x_{n+1}=x_k$.
The nontrivial case: $n+1\notin A_1\cup\cdots\cup A_n$.
Let $X=\mathbb N\setminus(A_1\cup\cdots\cup A_n)$, and note that the family $\mathcal C'=\{A\cap X:A\in\mathcal C\setminus[\mathcal C(x_1)\cup\cdots\cup\mathcal C(x_n)]$ is a linear cover of $X$.
By the lemma we can choose $A_{n+1}\in\mathcal C(n+1)\setminus[\mathcal C(x_1)\cup\cdots\cup\mathcal C(x_n)]$ and $x_{n+1}\in A_{n+1}\cap X$ so that, for all $z\in X\setminus A_{n+1}=\mathbb N\setminus(A_1\cup\cdots\cup A_n\cup A_{n+1})$ we have $\mathcal C'(z)\not\subseteq\mathcal C'(x_{n+1})$, that is, $\mathcal C(z)\not\subseteq[\mathcal C(x_1)\cup\cdots\cup\mathcal C(x_n)\cup\mathcal C(x_{n+1})]$. Now the conditions (i)–(iii) are satisfied for $i,j\le n+1$.
Corollary. If $\mathcal C$ is a linear cover of $X$ and if $\bigcup\mathcal C(x)$ is countable for each $x\in X$, then $\mathcal C$ has a minimal subcover.
Proof. The connected components of the hypergraph $(X,\mathcal C)$ are countable, so the theorem can be applied to each component separately.