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The result is not true. Take $A=\{\pi+n+1\colon n\in\mathbb N\}$, and $ B=\left \{-n-1+\frac1 {n+2}\colon n\in \mathbb N\right\} $. Both sets are closed, $\pi $ is a limit point of their sum, but is not on their sum.
If both $ A, B $ are compact, so is their sum (Since $ A+B $ is the image of $ A\times B $ under the continuous function $(x, y)\mapsto x+y $). Can you see what happens when one set is compact and the other isn’t?
Let $A$ be the set of negative integers.
Let $B$ be the set of all $n+\frac{1}{2^n}$ where $n$ ranges over the positive integers.
Then $A$ and $B$ are closed.
But $A+B$ is not closed, since it contains numbers arbitrarily close to $0$ but does not contain $0$.