For closed sets, is $\text{cl}(A+B)=\text{cl}(\text{cl}(A)+\text{cl}(B))$? Let $A$ and $B$ be nonempty subsets of $\mathbb{R}^n$, then is $\text{cl}(A+B)$ equal to $\text{cl}(\text{cl}(A)+\text{cl}(B))$? 
If that is true, then how to prove it? If they are not equal, then can you give me an example?
 A: Since $A+B$ is a subset of $\text{cl}A+\text{cl}B$, the inclusion $\subseteq$ is trivial.
For the other inclusion note that $\text{cl}A+\text{cl}B\subseteq \text{cl}(A+B)$ (see my answer here). Now apply the closure on both sides.
A: Suppose $x \in cl( cl(A) +  cl(B))$. Approximate $x$ by a sum $\overline a + \overline b$ where $\overline a \in cl(A)$ and $\overline b \in cl(B)$. Now approximate $\overline a$ by $a \in A$ and approximate $\overline b$ by $b \in B$. Using the triangle inequality, you'll find that $a+b$ approximates $x$. You ought to be able to conclude that $x \in cl(A+B)$ which gives the containment 
$$cl(cl(A) + cl(B)) \subseteq cl(A + B).$$
Are you able to fill in the details yourself?
A: The answer to your question in general is NO: For an example take $A=\lbrace (x,y)\in \mathbb R^2: x\ge 0, y\le f(x)\rbrace$ where $f(x)=\frac{x}{1+x}$ and $B=\lbrace (x,y)\in \mathbb R^2: (-x,y)\in A\rbrace$. These sets are closed. Drawing a picture you will see that $(0,2)$ is in the closure of the sum (analytically, $A+B \ni (n,f(n)) + (-n, f(n)) \to (0,2)$ in $\mathbb R^2$) but $(0,2)$ is not in $A+B$ since the second component of every element of $A+B$ is strictly smaller than $2$.

The (Minkowski) sum of two closed sets is again closed if one of the sets is compact.

As Carsten points out this answers a different question.
