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 $E$.