Tl;dr
A graph with this property (let's call it property P) must have uncountably many edges (for an example, see the answer of Florian Lehner).
The idea is to apply Zorn's lemma, and for that, we define a partial order on the set of all coverings:
$$C\ge\bar C\quad:\Longleftrightarrow\quad \mathrm m(C)\subset\mathrm m(\bar C).$$
Property P basically states that there is no maximal element. I will show the contrary via Zorn's lemma. Let $\mathfrak C$ be a chain. The tricky part is to construct an upper bound to that chain (note: we cannot just intersect all the $C\in\mathfrak C$, they might be disjoint).
If we assume that $E$ is countable, we can choose an enumeration $E=\{e_1,e_2,e_3,...\}$, and define the initial segments
$$E_i:=\{e_1,...,e_i\}.$$
I need the following term
Definition. Given a chain $\mathfrak C$, a subset $\mathfrak D\subseteq \mathfrak C$ is called end-dense, if for any $C\in \mathfrak C$ there is a $D\in \mathfrak D$ with $D \ge C$.
Note, that being end-dense is transitive.
We now recursively define a decreasing sequence of chains $\mathfrak C = \mathfrak C_0\supseteq \mathfrak C_1 \supseteq\cdots$, so that each $\mathfrak C_j$ is end-dense in $\mathfrak C_{j-1}$. Note that every covering $C\in \mathfrak C_{j-1}$ induces a covering on $E_j$, and there can be only finitely many distinct coverings on $E_j$. Consequently, we can choose an end-dense $\mathfrak C_j\subseteq \mathfrak C_{j-1}$ so that all $C\in \mathfrak C_j$ induce the same covering $\smash{\bar C_j}$ on $E_j$.
We then have $\bar C_1\subseteq \bar C_2\subseteq \bar C_3\subseteq \cdots$ and we can define $$\bar C := \bigcup_i \bar C_i.$$
I claim, that this is an upper bound for $\mathfrak C$: if a vertex $v$ is multiply covered by $\smash{\bar C}$, then so it is in some $\smash{\bar C_j}$. This $\bar C_j$ is induced by the infinitely many converings in $\mathfrak C_j\subseteq \mathfrak C$, hence $v\in\mathrm m(C),C\in \mathfrak C_j$. Since $\mathfrak C_j$ is end-dense in $\mathfrak C$, we obtain that $v$ is multiply covered by all $C\in \mathfrak C$.
Consequently, $\bar C$ is an upper bound for $\mathfrak C$, and Zorn's lemma establishes the existence of a maximal element in contradiction to your property P.