Does there exist $A \subset \mathbb{R}$, satisfying the following three conditions:
1)$\forall x \in A$ $\mu(A \cap [x - 1; x + 1]) > 1$
2)$\forall x \notin A$ $\mu(A \cap [x - 1; x + 1]) < 1$
3)$A \notin \{\mathbb{R}, \emptyset\}$
$\mu$ stands here for Lebesgue measure.
It is not hard to see, that if $A$ satisfies these conditions, then $\mathbb{R} \setminus A$, $-A$ and $x + A$, where $x$ is an arbitrary real number, satisfy these conditions too. Also, if $A$ and $B$ both satisfy these conditions and $\inf |A -B| \geq 2$, then $A \cup B$ satisfies those conditions too. However, that does not seem to be very helpful.
Another not very helpful fact is, that if $A$ satisfies these conditions and $(x ; x+1) \subset A$, then $\mu((x-1; x) \cap A) > 0$. And it again tells us nothing about existence of such subsets.
No, it does not.
$x \mapsto \mu(A\cap [x - 1; x + 1])$ is continuous on $\mathbb{R}$ as $\forall \epsilon > 0$ and $\forall x \in \mathbb{R}$ we have $|\mu(A\cap [x - 1; x + 1]) - \mu(A\cap [x + \epsilon - 1; x + \epsilon + 1])| \leq 2\epsilon$. Thus, by intermediate value theorem if $A \notin \{\mathbb{R}, \emptyset\}$, then $\exists x \in \mathbb{R}$, such that $\mu(A\cap [x - 1; x + 1]) = 1$