= Solution
Write $\ell(x,u)=t_+(x,u)-t_-(x,u)$ and let
$$
a(x)=\frac1{2\pi}\int_0^{2\pi}\mathbf1_{\{\ell(x,u_\theta)>0\}}\,d\theta>0.
$$
This is the accepted fraction of directions when zero-length chords are redrawn; it equals one in the interior. The <Markov kernel> integrates a test function $h$ as
$$
Ph(x)=\frac1{2\pi a(x)}\int_{\ell(x,u_\theta)>0}\frac1{\ell(x,u_\theta)}
\int_{t_-}^{t_+}h(x+tu_\theta)\,dt\,d\theta.
$$
The signed polar representation has <Jacobian determinant> $|t|$. Each point $y\ne x$ is represented twice, once with positive $t$ and once with negative $t$ and the opposite direction. Since the chord is the same in either representation, the <hit-and-run kernel density> is
$$
\boxed{p(x,y)=\frac1{\pi a(x)\ell(x,(y-x)/|y-x|)|y-x|}\quad(y\ne x)}
$$
for almost every $y\in\mathcal C$, and it is zero outside the body. Thus the transition is absolutely continuous with respect to planar <Lebesgue measure>. Redrawing removes the only potential zero-chord atoms.
Let $D=\operatorname{diam}(\mathcal C)>0$. For distinct $x,y\in\mathcal C$, the chord contains their segment, so it has positive length, and
$$
a(x)\leq1,\qquad\ell(x,u)\leq D,\qquad |y-x|\leq D.
$$
Hence $p(x,y)\geq1/(\pi D^2)$. Values on the diagonal, and on other null sets where a density is only specified almost everywhere, can be assigned this same lower bound. This gives a version satisfying
$$
\boxed{\inf_{x,y\in\mathcal C}p(x,y)\geq\frac1{\pi D^2}>0.}
$$
For interior $x,y$, $a(x)=a(y)=1$ and the chord length is unchanged on interchanging its endpoints, so $p(x,y)=p(y,x)$. A convex body's boundary has area zero; for example contractions about an interior point fill the interior and their areas tend to the body's area. Thus this symmetry holds almost everywhere for two uniform points, proving <detailed balance> and stationarity of the <uniform distribution>. Without positive area or the boundary convention in part (a), the literal assertion need not hold.
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