My friend and I have been working on trying to prove this inequality for a while. However, I think there is some trick we are just not seeing.
Suppose $F$ is a closed set in $\mathbb{R}$, whose complement has finite measure, and let
$$ \delta(x)=d(x,F)=\inf \{ |x - y| : y \in F \}.$$
Prove $$ \mid \delta(x) - \delta(y)| \le |x-y|.$$
Let $x,y\in \Bbb R$.
For all $z\in F$ we have $$\delta(x)\leq |x-z|\leq |x-y|+|y-z|,$$ hence $\delta(x)-|x-y|\leq |y-z|$ for all $z\in F$. It follows that $$ \delta(x)-|x-y|\leq \inf_{z\in F}|y-z|=\delta(y). $$
Now we get $\delta(x)-\delta(y)\leq |x-y|$.
By the same argument we can show also that $\delta(y)-\delta(x)\leq |x-y|$, so $|\delta(x)-\delta(y)|\leq |x-y|$.
Remark: The hypotheses on $F$ are unnecessary, the result holds for arbitrary subsets $F$, and an essentially identical argument shows that the result holds for arbitrary subsets $F$ of metric spaces $(X,d)$, where $\delta$ is defined by $\delta(x) = \inf\{d(x,y) : y\in F\}$.