An example for a graph with this property (let's call it property P) must have uncountably many edges.
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 $(C_i)_{i\in I}$ be a chain indexed by some ordered set $I$. The tricky part is to construct an upper bound to that chain (note: we cannot just intersect all the $C_i$, 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. A subset $J\subseteq I$ of an ordered set $I$ is end-dense, if for any $i\in I$ there is a $j\in J$ with $j\ge i$.
We now recursively define a decreasing sequence of ordered sets $I=I_0\supseteq I_1\supseteq\cdots$, so that each $I_j$ is end-dense in $I_{j-1}$. Note that every covering $C_i,i\in I_{j-1}$ 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 $I_j\subseteq I_{j-1}$ so that all $C_i,i\in I_j$ 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 every $C_i,i\in I$: 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 infinitely many of the coverings $C_i$, more precisely, by $C_i,i\in I_j$, where $I_j$ is end-dense in $I$. Hence $v\in\mathrm m(C_i),i\in I_j$. And since $I_j$ is end-dense, we obtain that $v$ is multiply covered by all $C_i$.
Consequently, $\bar C$ is an upper bound for the $C_i$, and Zorn's lemma establishes the existence of a maximal element in contradiction to your property P.