To show sum of two sets is closed in R^2 Let $A=\{(x,y)\in \mathbb R^2:\max\{|y|,|x|\}\leq 1\}$ and $B=\{(0,y)\in \mathbb R^2:y\in \mathbb R\}$. Show that $A+B$ is a closed subset of $\mathbb R^2$
My try:
let $z_n=x_n+y_n$ be a sequence in $A+B$ converging to $z\in \mathbb R^2$ .To show $z\in A+B$. Now $x_n\in A$
$\implies x_n=(a_n,b_n) $
where $|a_n|,|b_n|\leq 1$and $y_n=(0,c_n)$. Can you help me to proceed further please

Despite the fact that there is a more general result mentioned in another answer, in this specific case it might be easier to show that $$A+B=\{(x,y)\in\mathbb R^2; |x|\le 1\}$$ and then show that this set is closed.
If you try to draw a picture, you should see almost immediately what $A+B$ is, but let us try to do also a formal proof.
Let $C=\{(x,y)\in\mathbb R^2; |x|\le 1\}$. It is easy to see $A+B\subseteq C$.
On the other hand if $(x,y)$ is arbitrary point from $C$, i.e., if $|x|\le 1$, you wan to express it as the sum
$$(x,y)=(x,y_1)+(0,y_2)$$
where $|y_1|\le1$ and $y_2$ is arbitrary. Finding such $y_{1,2}$ should be rather straightforward, too.
