Minkowski's Convex Body Theorem

Consider the half-sized body $\frac{1}{2} K = \{\, x / 2 : x \in K \,\}$. Scaling in $d$ dimensions multiplies volume by $2^{-d}$, so the hypothesis $\operatorname{vol}(K) > 2^d \det(\Lambda)$ becomes $\operatorname{vol}(\frac{1}{2} K) > \det(\Lambda)$.

Let $F$ be a fundamental parallelepiped of $\Lambda$, so $\mathbb{R}^d = \bigsqcup_{v \in \Lambda} (F + v)$ and $\operatorname{vol}(F) = \det(\Lambda)$. Cut $\frac{1}{2} K$ into pieces $A_v = (\frac{1}{2} K) \cap (F + v)$ and translate each piece back into $F$ via $B_v = A_v - v \subseteq F$. Since the sum of the volumes of the $B_v$ equals $\operatorname{vol}(\frac{1}{2} K) > \operatorname{vol}(F)$, the sets $B_v$ cannot be pairwise disjoint (Blichfeldt's principle).

Pick distinct lattice vectors $u \neq v$ in $\Lambda$ with $B_u \cap B_v \neq \emptyset$. Then there exist two distinct points $p, q \in \frac{1}{2} K$ such that $p - u = q - v$, so $p - q = u - v \in \Lambda \setminus \{0\}$. Because $p, q \in \frac{1}{2} K$, we have $2p, 2q \in K$; central symmetry of $K$ gives $-2q \in K$, and convexity of $K$ places the midpoint $\frac{1}{2}(2p) + \frac{1}{2}(-2q) = p - q$ inside $K$. Thus $p - q$ is a nonzero lattice point in $K \cap (\Lambda \setminus \{0\})$.