My comment reposted as an answer:

If the continuum hypothesis holds, then we can give a well order $\prec$ to $\mathbb{R}$ isomorphic to the first uncountable ordinal. And then for each $j\in[-1,1]$ we can let $U_j=\{i\in[-1,1];j\prec i\}$ (so $\mu(U_j)=1$ for all $j$). Then for any $k\in(0,1]$ and any nonempty $I\subseteq[-k,k]$, if $I$ is uncountable then $\bigcap_{i\in I}U_i$ is empty. If not, there is some $j\in[-k,k]\setminus I$, and $U_j^c$ is countable.

And now, a construction without CH:

Let $U_x = \{y: \frac{1}{2}(x - 1) \leq y \leq \frac{1}{2}(x +1)\}$ as in the question. Note that If $D$ is any dense subset of $[-1,1]$, then for any nonempty $I\subseteq D$ we have

$$\ \ \mu\left(\bigcap_{i \in I} U_i \cap \bigcap_{j \in D\smallsetminus I} U^\mathsf{c}_j \right) = 0. \tag{1}$$

This is clear if $I=D$, and if not consider some point $x\in\overline{I}\cap\overline{D\setminus I}$ and note that when $i,j$ are close to $x$, $U_i\cap U_j^c$ is small.

So we can consider a bijection $f:[-1,1]\to[-1,1]$ such that $f([-1/n,1/n])$ is dense in $[-1,1]$ for all $n$ (see below for a construction), and define $V_x=U_{f(x)}$ for all $x\in[-1,1]$. Then the collection of sets $(V_x)_{x\in[-1,1]}$ satisfies what you want.

To construct the bijection $f:[-1,1]\to[-1,1]$, let $(A_n)_n$, $(B_n)_n$ be pairwise disjoint sequences of countable sets (the sets $A_i$ are pairwise disjoint, the sets $B_i$ are pairwise disjoint and the $A_i$ are disjoint with the $B_i$), such that $A_n\subseteq[-1/n,1/n]$ and $B_n$ is dense in $[-1,1]$ for all $n$.

Finally, define $f$ as the identity in $[-1,1]\setminus\bigcup_n(A_n\cup B_n)$, and let $f(A_n)=B_n$ and $f(B_n)=A_n$ for all $n$, using bijections between $A_n$ and $B_n$.

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