Let $X$ be an infinite set, and let ${\cal A}\subseteq{\cal P}(X)$ be a family of non-empty sets. We say $S\subseteq X$ is a cover for ${\cal A}$ if $A\cap S \neq \emptyset$ for all $A\in{\cal A}$.
Suppose that we have $A\neq B \in {\cal A}$ implies $|A\cap B|\leq 1$. Is there a cover $M\subseteq X$ for ${\cal A}$ such that for all $m\in M$ the set $M\setminus \{m\}$ is not a cover of ${\cal A}$?
I think I can do one case. Assume that $A_i\subseteq S$, $|A_i|=\aleph_0$ for $i\in I$, and $|A_i\cap A_j|\leq 1$ for $i\neq j$. By a theorem in the paper P. Komjáth: Families close to disjoint ones, Acta Math. Hung. 43 (1984), 199–204, there are cofinite sets $A'_i\subseteq A_i$ such that $\{A'_i:i\in I\}$ is disjoint.
Let $X$ be the following graph on $I$: $\{i,j\}\in X$ iff $(A_i-A'_i)\cap A'_j\neq\emptyset$. Given $i$ there is just finitely many $j$ as above, so there is an orientation of $X$ with all vertices having finite outdegrees. By an old result of Erdos and Hajnal, there is a well ordering $\prec$ of $I$ s. t. each vertex has only finitely many edges of $X$ going down. By transfinite recursion on $\prec$ pick $x_i\in A'_i$ for some $i\in I$ (let $I'$ be the set of those elements of $I$) as follows. If there is $j\prec i$ with $x_j\in A_i$, then set $i\notin I'$. Otherwise, pick $x_i\in A'_i-\bigcup\{A_j-A'_j:j\prec i\}$. This is possible, as the subtracted set is finite by the property of $\prec$. Now $B=\{x_i:i\in I'\}$ is obviously covering. Remove one element of it: $B'=B-\{x_i\}$ for some $i\in I'$. The way of the construction, $x_j\notin A_i$ for $j\prec i$. But there is no $i\prec j$ with $x_j\in A_i-A'_i$ either by the way $x_j$ was chosen, so $B'\cap A_i=\emptyset$.
We could not solve the original problem, but we could handle some more cases of the problem in 1.
1 T. Csernák, L. Soukup, Minimal vertex covers in infinite hypergraphs, Discrete Mathematics, Volume 346, Issue 4, April 2023.
A hypergraph possesses property $C({k},{\rho})$ iff $|\bigcap \mathcal E'|<{\rho}$ for each $k$ element set $\mathcal E'$ of hyperedges.
Komjáth proved that every uniform hypergraph possessing property $C(2,r)$ for some $r\in {\omega}$ has a minimal vertex cover. In this paper we relaxed the assumption of uniformity to an assumption that the set of cardinalities of the hyperedges is a ``small'' set of infinite cardinals, e.g. it is countable, or it does not contain uncountably many limit cardinals.
Komjáth also proved that GCH does not decide the following statement: "If a hypergraph $G$ possessing property $C({2},{{\omega}})$ is ${\mu}$-uniform for some ${\mu}\ge {\omega}_1$, then $G$ has a minimal vertex cover."
Using Shelah's Revised GCH theorem, we show that if we strengthen the assumption ${\mu}\ge {\omega}_1$ to ${\mu}\ge \beth_{\omega}$, then we can prove the statement in ZFC!
We also show that if all the hyperedges of a hypergraph are countably infinite,
then
instead of $C({2},{r})$ the assumption
$C({k},{r})$ (for some $k\in {\omega}$) is enough
to guarantee the existence of a minimal vertex cover.
If every hyperedge has cardinality ${\omega}_1$, then we can only prove that
$C({3},{r})$ is enough.
[Incorrect answer, perhaps mendable, perhaps not, sorry, see my comments below]
The answer is yes, the statement is in fact equivalent to AC.
AC is equivalent to "every set of pairwise disjoint nonempty sets has a choice set". So to see that the statement implies AC we can let ${\cal A}\subseteq{\cal P}(X)$ consist of pairwise disjoint sets.
Let $M\subseteq X$ be a minimal cover for ${\cal A}$ , that is: for all $m\in M$ the set $M\setminus \{m\}$ is not a cover of ${\cal A}$. Then $M$ is a choice set for ${\cal A}$. $$ $$ To see that AC implies the statement, we use Zorn's lemma.
For given ${\cal A}\subseteq{\cal P}(X)$ with the property that $A\neq B \in {\cal A}$ implies $|A\cap B|\leq 1$, let $H\subset {\cal P}({\cal A})\times{\cal P}(X)$ be defined by: $$ $$ $$H := \ \{(K,L)\in {\cal P}({\cal A})\times{\cal P}(X)\ |\ L\ \mbox{is a minimal cover for}\ K\ \}$$
$$ $$ We define a partial order on $H$ using set-inclusion: $$(K,L)\leq(K',L'):= K\subseteq K' \wedge L\subseteq L'$$ $$ $$ For $J=(K,L)\in H$ let $J_0=K, J_1=L$ be the coordinate projections.
We verify that every ascending chain in $(H,<)$ has an upper bound. Let $G\subseteq H$ such that $<$ is a total order on $G$. Then the upper bound $T$ for $G$ is given by:
$$T = (\bigcup_{J\in G} J_0, \bigcup_{J\in G} J_1)$$
It is straightforward to see that $\bigcup_{J\in G} J_1$ is again a minimal cover for $\bigcup_{J\in G} J_0$, so indeed $T$ is in $H$.
It is easy to see that there are non-empty ascending chains. By Zorn's lemma, $(H,<)$ contains a maximal element $U=(V,W)$. We claim that $V={\cal A}$, or in other words: $W$ is the desired minimal cover for ${\cal A}$.
To see that $V={\cal A}$, suppose there is $X\in{\cal A}, X\not\in V$.
case 1) $X\cap W \not= \emptyset$. Then $(V\cup\{X\},W)$ is larger than $(V,W)$, contradiction.
case 2) $X\cap W = \emptyset$. Pick $x\in X$. Then $(V\cup\{X\},W\cup\{x\})$ is larger than $(V,W)$, again contradiction.
