In a topological vector space, if $A$ is compact and $B$ is closed, then $A+B$ is closed. If $x\notin A+B$, then there is a neighborhood $V$ of $0$ such that $(x+V)\cap (A+B)=\emptyset$. But I am stuck in this step.
I have seen some solutions concerning about $\mathbb R^n$, but I don't know how to apply it to topological vector space which is more general.
 A: Assume that $x\in\overline{A+B}$ but $x\notin A+B$, then $(-A+x)\cap B=\emptyset$. So $-A+x\subseteq B^{c}$ and note that $B^{c}$ is open and $-A+x$ is compact.
Assume at the moment that there is some neighborhood $V$ of $0$ such that $(-A+x)+V\subseteq B^{c}$, then $((-A+x)+V)\cap B=\emptyset$.
Note that $((-A+x)+V)\cap B=(x+V)\cap(A+B)$.
On the other hand, since $x\in\overline{A+B}$, we must have $(x+V)\cap(A+B)\ne\emptyset$, so we have a contradiction.
Now it is left to show the fact assumed.
We are to show that, for compact set $K$ and open set $G$ such that $K\subseteq G$, there is a neighborhood $V$ of $0$ such that $K+V\subseteq G$.
For every $x\in K$, there is a neighborhood $V_{x}$ of $0$ such that $x+V_{x}\subseteq G$. Since the addition is continuous, there is a neighborhood $W_{x}$ of $0$ such that $W_{x}+W_{x}\subseteq V_{x}$.
So $\{x+W_{x}\}_{x\in K}$ is an open cover for $K$ and hence there is some finite subcover $x_{i}+W_{i}$, $i=1,...,n$ for $K$. Let $V=\displaystyle\bigcap_{i=1}^{n}W_{i}$, then $V$ is also a neighborhood of $0$.
For $u\in K+V$, then $u=x+v$ for some $x\in K$ and $v\in V$, and we have some $i$ such that $x\in x_{i}+W_{i}$ and hence $u\in x_{i}+W_{i}+V\subseteq x_{i}+W_{i}+W_{i}\subseteq x_{i}+V_{x_{i}}\subseteq G$, this shows that $K+V\subseteq G$.
