Distance function to $\Omega\subset\mathbb{R}^n$ differentiable at $y\notin\Omega$ implies $\exists$ unique closest point I am trying to show the following two statements are true: 
(1) For any nonempty set $\Omega\subset\mathbb{R}^n$, the set $B$ consisting of points $y\notin\Omega$ where there is not a unique closest point $x\in\partial\Omega$ for each $y$ has $\mathcal{H}^n$-measure 0. 
(2) This implies that the number of points $a\in A\subset \mathbb{R}^2$ such that  $\mathcal{H}^{1}(\vec{n}(a)\cap B)>0$ is countable, where $\vec{n}(a)$ is the normal to $a\in A$ and $A$ is $C^1$ (i.e. it is locally the graph of a $C^1$ function from $\mathbb{R}$ to $\mathbb{R}$. More specifically, $A$ is a $C^1$ closed embedded 1-d submanifold in $\mathbb{R}^2$ and is thus a compact set in $\mathbb{R}^2$.). 
For (1), I would like help proving only this related fact: if the distance function $f$ to $\Omega$ is differentiable at a point $y\notin\Omega$, then there exists a unique closest point $x\in\partial\Omega$ to $y$.
I thought that I could get at this by saying that the graph of $f$ has a sharp corner wherever there is more than one closest point (and hence is not differentiable there). But, I am told this is not always true. 
For (2), I thought that, since we're in $\mathbb{R}^2$, if we have an uncountable number of segments with positive $\mathcal{H}^{1}$ measure, then $B$ would have to have positive $\mathcal{H}^{2}$ measure since $B$ contains these segments. I'm told this is not true either. Please assist. 
 A: Just a partial answer i.e. a longer hint for (1):
Let us call the distance function $d(x):=\inf\{|x-a|,\ a\in\Omega\}$. If $d$ is differentiable in $x\in\mathbb{R}^n\setminus\overline{\Omega}$, you can show, that
$$a_0:=x-d(x)\nabla d(x)$$
is the unique closest point in $\partial\Omega$ to $x$. To show this you should first prove the following: Let $\overline{a}\in\partial\Omega$, such that $|x-\overline{a}|=d(x)$. Then for every $0\leq t\leq 1$ you have
$$d((1-t)\overline{a} + tx)=t|x-\overline{a}|.$$
If you have done that you should calculate
$$\frac{\partial}{\partial t}( d((1-t)\overline{a} + tx))=\frac{\partial}{\partial t}(t|x-\overline{a}|)$$
at $t=1$ and by chain rule. Afterwards use Cauchy-Schwartz inequality and $|\nabla d(x)|\leq 1$ ($d$ is Lipschitz with Lipschitz constant $\leq 1$) to show $x-\overline{a}=\lambda \nabla d(x)$ with $\lambda=\frac{1}{d(x)}$. You can insert this into the definition of $a_0$ to show that $a_0=\overline{a}$.
A: For the uniqueness:
You have $|d(y_1, \Omega)-d(y_2, \Omega) |\le \|y_1-y_2\|$. Moreover, for $y\not \in \bar \Omega$ and $y^{\star}$ the closest point in $\bar \Omega$ to $y$ at distance $\delta$ you have 
$$d(y+ t\frac{(y^{\star}- y)}{\|y^{\star}-y\|}, \Omega) = (\delta-t)$$
for any $t\in [0, \delta]$. Therefore, if $d_{\Omega} =d(\cdot, \Omega)$ is differentiable at $y$ then the gradient $\nabla d_{\Omega}(y)$ is $-\frac{(y^{\star}- y)}{\|y^{\star}-y\|}$. Uniquenss should follow. 
