Proof. Let $D$ be any fundamental parallelepiped. Then obviously$$\Reals^n = \coprod_{x\in\Lat} (D+x$$ (where $\coprod$ means disjoint union) and thus$$\frac{1}{2}\Reg = \coprod_{x\in\Lat} \left(\frac{1}{2}\Reg\cap(D+x)\right)$$ Now, note that$$\frac{1}{2}\Reg\cap (D+x)=\left(\left(\frac{1}{2}\Reg-x\right)\cap D\right)-$$ (draw a picture!) and thus, since measure is preserved by translation,$$\mu\left(\frac{1}{2}\Reg\cap (D+x)\right)=\mu\left(\left(\frac{1}{2}\Reg-x\right)\cap D\right$$ so that if all the $\frac{1}{2}\Reg-x$ are disjoint, we have$$2^{-n}\mu(\Reg)=\mu\left(\frac{1}{2}\Reg\right) = \mu\left(\coprod_{x\in\Lat} \left(\frac{1}{2}\Reg\cap(D+x)\right)\right)=\sum_{x\in\Lat}\mu\left(\left(\frac{1}{2}\Reg-x\right)\cap D\right)\leq \mu(D) = \Delt$$ which is a contradiction. Thus there must exist $x\neq y\in\Lat$ and $c_1,c_2\in \Reg$ such that$$\frac{1}{2}c_1-x = \frac{1}{2}c_2-y$$ Thus $x-y=\frac{1}{2}(c_2-c_1)\in \Reg$ since $\Reg$ is convex and centrally symmetric, and certainly $x-y\in\Lat$ , so we have found a nonzero element of $\Reg\cap\Lambda$ .

Proof. Apply the previous case to $C_n=\left(1+\frac{1}{n}\right)C$ , i.e. dilate $C$ . This gives a sequence of points $x_1, x_2, \ldots, x_n, \ldots$ with $x_i\in \Lambda\cap C_i-\{0\}$ . But $\Lambda$ is discrete, so there must be a subsequence constant at a nonzero element$$x \in \Lambda \bigcap \left( \bigcap_{i=1}^{\infty} C_i-\{0\} \right) = \Lambda \cap \overline{C}-\{0\}$$ Since $C$ is compact and thus closed, $x\in C$ .