Nearest point function from sphere to convex set is surjective onto the boundary of the convex set Let $A$ be a bounded convex subset of $\mathbb{R}^n$ and $f: \mathbb{R}^n \to \mathbb{R}^n$ such that $f(x)$ is is the nearest point in $A$ to $x$, which is unique because $A$ is convex. I know that $f$ is a contraction (and therefore continuous). I am interested in proving there exists a sphere $S$ containing $A$ (i.e. $A \subset B$ and $\partial B = S$, where $B$ is a ball) such that $\partial A \subset f(S)$. 
I have the general idea that there must be a hyperplane containing a point $x \in \partial A$ such that there are only elements of $\partial A$ on one side of the hyperplane, in which case the point $y$ where the line normal to the hyperplane at $x$ intersects $S$ has $f(y) = x$, but I don't know how to prove any of that.
I am using this to prove that $A$ is rectifiable (by proving $\partial A$ is measure zero. I have already proven the continuous image of a measure zero set is measure zero, so we would have $\partial A \subset f(A)$, which is measure zero and thus $\partial A$ is measure zero), so if there is a better way to do that I would be happy to hear. I also don't really know what to tag this, any input would be welcome. I don't have any background in measure theory though, so I would appreciate answers which didn't involve much measure theory.
The definition I was given for a measure zero set is that $A$ is measure zero if for any $\varepsilon > 0$ there exists a countable collection of rectangles $Q_i$ such that $S \subset \bigcup_i Q_i$ and $\sum_i \text{vol}(Q_i) < \varepsilon$.
 A: Take $x\in \partial A$. There exists a hyperplane $H$ such that contains $x$ and such that $A$ lies on one side of $H$. Let $n$ a normal vector to $H$. The equation of the hyperplane is $n \cdot (z-x)=0$. Now, for every 
$z\in A$ we have $n\cdot (z-x)\le 0$. Therefore, for every $y= x + t n$, with $t\ge 0$  and $z \in A$ we have $\|y-z\|\ge \|y-x\|$, with equality only if $z = x$ (draw a picture and notice an obtuse triangle). So for every point $y= x + t n$ the closest point in $A$ to $y$ is $x$.  Now given any sphere containing $A$, the ray $y= x + t n$ will intersect this sphere ( in exactly one point).  Therefore, for every sphere containing $A$ the map (closest point in $A$) from the sphere to the boundary of $A$ is surjective (it may not be injective, think of a square inside a circle). 
$\bf{Added:}$ We can generalize the statement as follows: Let $V$ a finite dimensional normed space with a strictly convex norm. Let $A$ a closed convex set. The map $y\mapsto \phi(y)$-- the closest point in $A$ to $y$-- is well defined on $V$ and continuous. If $y\not \in A$ then $x\colon =\phi(y)\in \partial A$. For every $y'$ in the ray $xy$ we have $\phi(y') = x$. This is easy to see if $y'$ is in the segment $[x,y]$. Otherwise, $y$ is in the segment $[x,y']$. Let $z'$ any point in $A$. Consider $z$ on the segment $[x,z']$ so that $y'z'$ and $yz$ are parallel. $z\in A$ since $A$ is convex. We have $\|y-z\|\ge \|y-x\|$ so $\|y'-z'\|\ge \|y'-x\|$. 
Consider now $S$ a compact set surrounding $A$ ( that means $A$ is contained in a bounded component of $V\backslash S$). Any ray $xy$ as above will intersect $S$. Therefore, $\phi(S) = \phi(V\backslash A)$. 
Let $x\in \partial A$ and  $y_n\in V\backslash A$, $y_n\to x$. We have $\phi(y_n) \to \phi(x) = x$. Hence $\phi(V\backslash A)$ is dense in $\partial A$. From above we conclude $\phi(S)$ is dense in $\partial A$. But $\phi(S)$ is a compact subset of $\partial A$, since $S$ is compact, and $\phi$ continuous. We get $\phi(S)=A$. 
