If $X$ and $Y$ are compact subsets of $\mathbb R^n$, then $X+Y$ is compact. 
If $X$ and $Y$ are compact subsets of $\mathbb R^n$, then $X+Y$ is compact.

Is the statement true in NLS.
"Since $X$ and $Y$ are compact then $X\times Y\subseteq\mathbb{R}^{2n}$ is compact, and use the continuity of the function $(x,y)\mapsto x+y$ and the fact that continuous images of compact sets are compact, to conclude that $X+Y$ is compact."
Can this idea be used?
 A: I think it's a great idea. I even think it's a full proof, unless you need to prove that $(x,y)\to x+y$ is continuous, or that the product of two compact spaces is compact.
A: An alternative approach is to use sequentially compactness. Since our space is metrizable, this is equivalent to compactness.
Let $(z_n)_{n \in \mathbb{N}}$ be any sequence in $X+Y$, i.e. we can write $z_n = x_n +y_n$. Taking a subsequence, we can guarantee that $x_{n_k} \rightarrow x$ for some $x \in X$. Now again there exists a subsequence $(n_{k_l})_{l \in \mathbb{N}}$ of $(n_k)_{k \in \mathbb{N}}$ such that also $y_{n_{k_l}} \rightarrow y$ for some $y \in Y$. Altogether, we see that $z_{n_{k_l}} \rightarrow x+y$ and $x+y \in X+Y$. (Since $|z_{n_{k_l}}-(x+y)| \leq |x_{n_{k_l}} -x| + |y_{n_{k_l}}-y| \rightarrow 0$. Of course, this shows implicitly that the map $(x,y) \mapsto x+y$ is continuous. So your argument is somewhat shorter.)
A: In any Abelian topological group, the sum of a closed set and a compact set is closed.
The following is a crucial fact to prove the above statement:

In any topological group (Abelian), if the compact set $K$ is contained in the open set $U$ then there is a neighborhood $V$ of $0$ such that $K+V\subseteq U$.

