(I) $c$ is convex curve if curve lies in one half plane wrt
tangent line. If $c$ is a simple closed convex curve then $c$
bounds a convex set $K$ : For $p,\ q\in K$, if $l$ is a line segment from
$p$ to $q$ and if $l\subset K$, then we are done\
If not, there exists $r\in l$ s.t. $r$ is not in $K$. Then WLOG we have $$
r',\ r''\in l\cap \partial K,\ pr'<pr''$$
and $ pr',\ qr''\subset K$ and $r'r'' -\{ r',r''\}$ is not in $K$
(1) If $l$ is not in $T_{r'}\partial K$, since $c$ is in one half
plane $P$ wrt $T_{r'} \partial K$ then $$ c\subset P,\ r''\in P $$
Hence $K\subset P$ so that it is contradiction to the fact that $p$
is not in $P$ (Here $l$ meets $T_{r' } \partial K$ transversally).
(2) If $l$ is in $T_{r'}\partial K$, note that arc $r'r''$ in $c$ is
not in $K$ Hence if $r$ is closed to $r'$ then $T_r\partial K$ meets
$T_{r'}\partial K$ transversally That is $T_r\partial K$ meets $l$
transversally
(II) Convex hull $C$ of $\Gamma$ is a smallest convex set containing
$\Gamma$ Note that since $K$ is convex so $C\subset K$ If $x\in
K-C$, then $x$ is interior point of $K$. Note that if $l_p$ is a
line segment from $x$ to $p\in \partial K$, then $l_p$ is in $K$ And
there exists $-p\in \partial K$ s.t. $l_p\cap l_{-p}$ is line
segment. Hence $x\in p(-p)$ So $x\in C$.
Proof of $D=K $ : Since $K$ is convex hull and $D$ is convex containing $\Gamma$, so
$K\subset D$. If $x\in D-K$, there exists $h\in K$ s.t. $$
d(h,x)=d(K,x) $$
Since $D$ is in a side of tangent line of $\gamma$ at $h$, so it is
contradiction.
Based on your description, I think I've found an argument that works. However many points there are, space them on a segment of helix defined by the following equation in cylindrical coordinates:
$$
(r, \theta, z) = (1, \theta, \frac{\theta}{2}) \hspace{1cm}\text{for } 0\leq\theta\leq\frac{\pi}{2}
$$
The relevant solids will be the Voronoi cells of these points, cut off at some large distance. We want to check the existence of a boundary between every chosen pair of points on our helix segment. It will suffice to show that for every pair of points on the helix (not just the ones we happened to choose), there will be some point $Q$ in space that is equidistant from the pair and is closer to that pair of points than to any other point on the helix. (The existence of such a point shows that there must be a boundary between the two corresponding cells, with $Q$ lying on that boundary.)
The general idea of the proof is that if you're standing a few steps behind the center of a circle and looking straight through the center and across to the opposite side, the points with a larger angle to your line of sight will be closer to you. On the other hand, if you're looking at a line, the points with a larger angle will be farther. In the helix, which is kind of a cross between a circle and a line, these effects can be balanced, and the nearest points to you will be at an intermediate angle to you line of sight.
Let $P_1 = (1, \theta_1, \theta_1/2)$ and $P_2 = (1, \theta_2, \theta_2/2)$. We'll want to find a point $Q$ in space that is closer to those two points than to any other point on the helix segment. We'll choose $Q$ to be the following, for some $a$, which we'll specify later:
$$(-a, \frac{\theta_1+\theta_2}{2}, \frac{\theta_1+\theta_2}{4})$$
Define $R(\phi)=(1, \phi+\frac{\theta_1+\theta_2}{2}, \frac{\phi}{2}+\frac{\theta_1+\theta_2}{4})$. (This is always a point on the helix.) Note that since the helix segment only spans an angle of $\pi/2$, we only need to worry about $\phi$ between $-\pi/2$ and $\pi/2$. Also, note that $R(\phi^*)=P_1$ and $R(-\phi^*)=P_2$ for $\phi^*=\frac{\theta_1-\theta_2}{2}$.
We can now compute the squared distance between $R(\phi)$ and $Q$:
$$
s^2 = \frac{\phi^2}{4} + (\sin\phi)^2 + (a+\cos\phi)^2 = \frac{\phi^2}{4} + 1 + a^2 + 2a\cos\phi
$$
Setting the derivative to 0, so we can find possible locations of minimal distance:
$$
\frac{ds^2}{d\phi} = \frac{\phi}{2} - 2a\sin\phi = 0
$$
This yields the solution $\phi=0$ for all $a$, and an additional pair of non-trivial solutions, $\phi = -\phi_a, \phi_a$ for $\frac{1}{4}<a<\frac{\pi}{4}$. (We're only counting solutions in $(-\frac{\pi}{2}, \frac{\pi}{2})$, which is why there's an upper limit on $a$ too.)
We have $\lim_{a\to \frac{1}{4}}\phi_a = 0$ and $\lim_{a\to \frac{\pi}{4}}\phi_a = \frac{\pi}{2}$, and $\phi_a$ varies continuously with $a$. So we can choose an $a$ to produce any desired value of $\phi_a$ between 0 and $\frac\pi2$.
Further, we can compute the second derivative of $s^2$, which is:
$$
\frac{d^2s^2}{d\phi^2} = \frac{1}{2} - 2a\cos\phi
$$
This is negative for $a>\frac{1}{4}$, indicating that the $\phi=0$ critical point is a local maxima. Therefore, the neighbouring $-\phi_a$ and $\phi_a$ solutions must be local minima, and since $s^2(\phi)$ is an even function, they must have the same value of $s^2$ for both. Also, since there are no other possible local minima for $\phi\in(-\frac{\pi}{2}, \frac{\pi}{2})$, $\phi_a$ and $-\phi_a$ must give the global minima on that interval. Therefore, for given values of $\theta_1$, $\theta_2$, we choose $a$ such that $\phi_a = \phi^* = \frac{\theta_1 - \theta_2}{2}$, ensuring that $P_1$ and $P_2$ will correspond to $\phi_a$ and $-\phi_a$ and will therefore be the closest points to $Q$.
Best Answer
Just a thought. Welcome to fill in the details.
First let's fix the ball radius as $r$. Let $$ f(x,r)=\frac{\mu(B(x,r)\cap C)}{\mu(B(x,r))} $$
This should be rather obvious but I am not sure how to give a formal proof.
Now you can fix any $r_0>0$ as a start and obtain $F(x)=f(x,r_0)$ as a continuous and positive function on $C$ by Lemma 1. Since $C$ is compact, $F(x)$ assumes a positive minimum value, say $\lambda(C)=\inf_{x\in C}F(x)>0$. Now choose $r_0$ as your $\epsilon$. Then for any $x\in C$, $r<\epsilon=r_0$, we have from Lemma 2: $$ f(x,r)\geq f(x,r_0)\geq\lambda(C) $$
Note that $\lambda(C)$ depends on $r_0$ and is only one lower bound but not necessarily the greatest one. I conjecture if you let $r_0\to0$, then $\lambda(C)$ will approach or even become the greatest lower bound.