 Solution

False. In $\mathbb{R}$, letEach is closed: its distinct points escape to infinity, so it has no finite accumulation point. Their Minkowski sum contains $1/n$, tending to zero. However zero is not in the sum. If $m-n+1/n=0$ with positive integers $m,n$, then $n-m=1/n$. For $n>1$ the right side is strictly between zero and one, while the left side is an integer; for $n=1$ the equation would require $m=0$. Thus the sum of two closed sets need not be closed. The same counterexample embeds along a coordinate axis in ${\mathbb{R}}^n$.