Covering null sets by a finite number of intervals

Let us say that a subset $A$ of $\mathbb R$ has property $P$ if for every $\epsilon>0$ there is a finite collection of open intervals $(a_1,b_1),(a_2,b_2),\cdots,(a_n,b_n)$ such that $A \subset \cup_i (a_i,b_i)$ and $\sum (b_i-a_i) <\epsilon$. My question: if $A$ is a nowhere dense set with measure $0$ does it have property $P$?

Some basics: every compact set of measure $0$ (obviously) has property $P$. No dense set of measure $0$ can have property $P$. More generally, if $A$ has property $P$ then $A$ is nowhere dense. Proof: if $(\alpha,\beta) \subset \overset {-} {A}$ take $\epsilon <\beta -\alpha$. If $A \subset \cup_i (a_i,b_i)$ and $\sum (b_i-a_i) <\epsilon$ then $(\alpha,\beta) \subset \cup_i [a_i,b_i]$ so $\beta -\alpha <\epsilon$, a contradiction. My question is if every nowhere dense set of measure $0$ has property $P$. My guess that the implication does not hold but I don't have a counterexample.

For an example of a bounded nowhere dense set of measure zero which does not have your property $P$, let $F$ be a compact nowhere dense set of positive measure (a fat Cantor set), and let $A$ be a countable dense subset of $F$.

In fact, it's easy to see that a set $A\subseteq\mathbb R$ has property $P$ if and only if the closure of $A$ is a compact set of measure zero.

The set $\mathbb Z$ has measure $0$ and it is nowhere dense. However, the property $P$ doesn't hold for $\mathbb Z$.