Let $V:=\omega\times\omega$ and $E=\{E_{n,m}:n,m\in\omega\}$ where $$E_{n,m}:=(\{0,\dots,n\}\times\{m\})\cup\{(2n,m+1)\}.$$ It seems that the hypergraph $(V,E)$ has no minimal cover.

A simplification of this example was suggested by Gerhard Paseman in his comment:
Just take $V=\mathbb Z$ and $E=\{E_n,F_n:n\in\mathbb N\}$ where $E_n:=\{-2n\}\cup [0,n]$ and $F_n:=[-n,0]\cup\{2n\}$.

**Added in Edit.** I have just discovered that the Paseman's example is "isomorphic" to the example of @domotorp (Strongly minimal covers) given in 2015 to a similar question of Dominic van der Zypen.

The question of Dominic van der Zypen has positive answer if the cardinalities of hyperedges are upper bounded by some number.

**Theorem.** For any $n\in\mathbb N$, any set $V$ and family $E\subset[V]^{\le n}$ with $\bigcup E=V$ there exists a minimal subfamily $C\subset E$ such that $\bigcup C=V$.

*Proof.* For $n=1$ the assertion is trivial. Assume that for some $n\ge 2$ we have proved that any family $E\subset [V]^{<n}$ with $\bigcup E=V$ contains a minimal subfamily $C\subset E$ with $\bigcup C=V$.

Take any family $E\subset [V]^{\le n}$ with $\bigcup E=V$. Using Zorn's lemma, choose a maximal disjoint subfamily $D\subset E$. If the set $W:=V\setminus\bigcup D$ is empty, then $D$ is a required minimal cover of $V$.

So, we assume that $W$ is not empty. By the maximality of $D$, each hyperedge $e\in E\setminus D$ intersects the set $\bigcup D$, which implies that $E':=\{e\cap W:e\in E\setminus D\}\subset [W]^{<n}$. By the inductive assumption, the family $E'$ contains a minimal subfamily $C'\subset E'$ with $\bigcup C'=W$.

For every $c\in C'$ find a hyperedge $e_c\in E\setminus D$ such that $e_c\cap W=c$. Let $H:=\{e_c:c\in C'\}$ and $D':=\{e\in D:e\not\subset \bigcup H\}$. It can be shown that $C:=H\cup D'$ is a minimal subcover of $E$.