Closed And Compact Sets For $A \subset \mathbb {R}, B \subset \mathbb {R}$.
Let $A + B := {a + b : a \in A, b \in B}$
Let $A$ be a closed set and $B$ be a compact set. Show that $A + B$ is closed
 A: I think you mixed up $A$ and $B$ in your proof. So let's assume that $A$ is compact and $B$ is closed. 
You do not need the second subsequence. $(x_n+y_n)$ converges, say towards $z$. $(x_{n_k})$ converges towards $x \in A$, hence $(y_{n_k})$ converges towards $z-x$. Let $y = z - x$. As $B$ is closed, $y \in B$. Proof is complete 
A: Of course, the goal is not to show that $(x_n + y_n)$ converges (it might not). What we need to do is
1) take a sequence $(x_n + y_n) \in A + B$ that converges to a point $z$.
2) show that $z \in A + B$
3) we should be able to do this by showing $x_n \to x$ and $y_n \to y$ (necessarily $x \in A$ and $y \in B$ since $A, B$ are closed)
4) this will show that $(x_n + y_n) \to z = x + y \in A + B$ (possibly we may need to pass to a subsequence)
It is easier to write down the proof if we give the name $z$ to what $x_n + y_n$ converges to so that once we use compactness to get a limit $y_{n_k} \to y$, we can talk about $z - y$
Specifically, we have
$$x_{n_k} = (x_{n_k} + y_{n_k}) - y_{n_k} \to z - y$$
If we call this $x$ then $x \in A$ and we have $z = x + y \in A + B$.
