**Delete reason:** I believe my argument is incomplete. I might work on it later.


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An example for this property must have uncountably many edges.
My argument might be extened to show that no example exists, but I feel not safe enough in handling transfinite induction arguments to do that myself.

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The idea is to apply [Zorn's lemma](https://en.wikipedia.org/wiki/Zorn%27s_lemma), and for that, we define a partial order on the set of all coverings:

$$C\ge\bar C\quad:\Longleftrightarrow\quad \mathrm m(C)\subset\mathrm m(\bar C).$$

Your property basically states that there is no maximal element. I will show the contrary via Zorn's lemma. Let $C_1\le C_2\le\cdots$ be a chain. The tricky part is (as always) 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 $e_1,e_2,...\in E$, and define 

$$E_i:=\{e_1,...,e_i\}.$$

There can be only finitely many distinct coverings on any $E_i$. We then define recursively sequences $C_1^{(j)},C_2^{(j)},...$ for $j=0,1,2,...$. For $j=0$, let this just be the sequence $C_1,C_2,...$. For $j \ge 1$, let this be a subsequence of $C_1^{(j-1)},C_2^{(j-1)},...$, so that all of these induced the same covering, say $\bar C_j$, on $E_j$. This is possible because of the pidgeon hole principle.

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 to $C_1,C_2,...$. Because if any vertex $v$ is multiply covered by $\bar C$, then so it is in some $\bar C_j$. This $\bar C_j$ is induced by infinitely many of the coverings $C_i$, and hence, infinitely many of these cover $v$ twice. The way the partial order is defined then ensures that $v$ is multiply covered by all coverings $C_i$.