For $r=\frac{1}{2}$, I have a geometric argument.

Let $I=[-\frac{1}{2},\frac{1}{2}]$.

Now $Vol(D^{n-1} \times I) = Vol(D^{n-1})$. Take an annulus $A$, say with outer radius $\frac{1}{2}$ and inner radius $0.9\frac{1}{2}$. Remove $A \times [0.9\frac{1}{2}, \frac{1}{2}]$ from the top of $D^{n-1} \times I$. This still contains $D^n$ and has volume at least 0.1% less than $Vol(D^{n-1} \times I) = Vol(D^{n-1})$. So $Vol(D^n) < 0.999 Vol(D^{n-1})$. Done.