Proof check of sum of a compact and closed set of real numbers is closed

Let $A$ be a closed and $B$ be a closed and bounded set in $\mathbb R$ , then we have to show that $A+B:=\{a+b:a\in A , b\in B \}$ is closed in $\mathbb R$ .

My Proof : Let $\{a_n+b_n\}$ be a convergent sequence in $A+B$ , where $\{a_n\}\in A , \{b_n\}\in B$ , with limit $x$ , we have to show that $x \in A+B$ . Since $B$ is closed-bounded , there is a subsequence $\{b_{r_n}\}$ of $\{b_n\}$ such that $\{b_{r_n}\}$ converges with $\lim \{b_{r_n}\}=l$ (say) ; then since $B$ is closed , $l\in B$. Also $\lim \{a_{r_n}+b_{r_n}\}=x$ , thus $\lim \{a_{r_n}\}=x-l$ , and since $A$ is closed , so $x-l \in A$ , thus $x=(x-l)+l \in A+B$ . Am I correct ?

In the above proof, things become easy if we take $z_n=a_n+b_n \in A+B $ . Let $\lim z_n=z \in \mathbb{R}$ Since $B$ is compact, $(b_n)$ has a convergent subsequence, let's say $(b_{n_r})$. Let $\lim b_{n_r} = l \in B $ . So $ a_{n_r}=z_{n_r}-b_{n_r} $. So $(a_{n_r})$ is also convergent and let $\lim a_{n_r}=m .$ As $A$ is closed, $m\in A$ . So $z=m+l \in A+B $ which completes the proof.