What is shown below is a reference from the text Functional Analysis by Walter Rudin.
If $X$ s a vector space over the field $\Bbb F$ then the following notation well be used $$ x+A=\{x+a:a\in A\}\\ x-A:=\{x-a:a\in A\}\\ A+B:=\{a+b:a\in A\wedge b\in B\}\\ \lambda A:=\{\lambda a:a\in A\} $$ for any $A,B\subseteq X$, for any $x\in\ X$ and for any $\lambda\in\Bbb F$.
Definition
If $\tau$ is a topology on a vector space $X$ such that
- every point of $X$ is a closed set
- the vector space operation are continuous with respect to $\tau$
we say that $X$ is a topological vector space equipped with the vector topology $\tau$.
So we note that the condition $1$ imply that any topological vector space will be $T_1$ but many authors omitted the above condition from the definition of a topological vector space.
Lemma
Let be $X$ a topological vector space over the field $F$. So if $W$ is a neighborhood of $0$ in $X$ then there is a neighborhood $U$ of $0$ which is symmetric (that is $U=-U$) and which satisfies $U+U\subseteq W$.
Proof. Omitted.
Now we observe that the proposition can applied to $U$ in place of $W$ so that there exist a new symmetric ceighborhood $U$ of $0$ such that $$ U+U+U+U\subseteq W $$ because if there exist a symmetric neighborhood $V$ of $0$ such that $V+V\subseteq U$ then $(V+V)\times (V+V)\subseteq U\times U$ and so $$ V+V+V+V=s\big((V+V)\times(V+V)\big)\subseteq s(U\times U)=U+U\subseteq W $$ where $s:X\times X\rightarrow X$ is the vector sum.
So I like to discuss some passages of the following theorem.
Theorem
Suppose $K$ and $C$ are disjoint subsets of a topological vector space X such that K is compact and C is closed. so there exist a symmetric neighborhood $V$ of $0$ such that $$ (K+V)\cap(C+K)=\emptyset $$
Proof. If $K$ is empty tehn $K+V$ is too and so the concludion follows immediately. We therefore assume that $K$ is not empty so that we consider a point $x$ of this set. So since $C$ is closed, since $x$ is not in $C$ and since the topology of $X$ is invriant under translations, the preceding proposition shows that $0$ has a symmetric neighborhood $V_x$ such that $x+V_x+V_x+V_x$ does not intersect $C$ and so the symmetry of $V_x$ show then that $$ (x+V_x+V_x)\cap(C+V_x)=\emptyset $$ Since $K$ is compact, there are finitely many points $x_1,...,x_n$ in $K$ such that $$ K\subseteq\bigcup_{i=1}^n\big(x_i+V_{x_i}\big) $$ so that if we put $$ V:=\bigcap_{i=1}^n V_{x_i} $$ then $$ K+V\subseteq\bigcup_{i=1}^n\big(x_i+V_{x_i}+V\big)\subseteq\bigcup_{i=1}^n\big(x_i+V_{x_i}+V_{x_i}\big) $$ and no term in this last union intersects $C+V$ by what above shown and so the theorem follows.
So unfortunately I did not understand why $\big(x+V_x+V_x+V_x\big)$ is disjoint from $C$ and then why this implies that $\big(x+V_x+V_x\big)$ and $\big(C+V_x\big)$ are disjoint. Then I like to discuss my arguments below to explain the inclusions of the theorem are correct. Anyway I observe that since $X\setminus C$ is open then $-x+(X\setminus C)$ is too and since $0$ is there then there exist a open symmetric neighborhood $V_x$ of $0$ such that $V_x+V_x\subseteq -x+(X\setminus C)$ and if $s$ is the vector sum this (neighborhood) is such that $$ x+V_x+V_x=s\big[\{x\}\times(V_x+V_x)\big]\subseteq s\Big[\{x\}\times\big(-x+(X\setminus C)\big)\Big]=X\setminus C $$ where $s$ is the vector sum but this does not mean that $x+V_x+V_x+V_x$ is disjoint from $C$. Now if $0\in V_x$ for each $x\in K$ then clearly $x\in x+V_x$ which is open and so the collection $\mathfrak V:=\{x+V_x:x\in K\}$ is an open cover of $K$ by which the existece of $x_1,...,x_n$ follows for the compactness of $K$. So by the first inclusion we conclude that $$ K+V=s[K\times V]\subseteq s\Big[\big(\bigcup_{i=1}^n x_i+V_{x_i}\big)\times V\Big]=s\Big[\bigcup_{i=1}^n\big((x_i+V_{x_i})\times V\Big]=\bigcup_{i=1}^n s[(x_i+V_{x_i})\times V]=\bigcup_{i=1}^n(x_i+V_{x_i}+V) $$ and finally if $V\subseteq V_{x_i}$ for each $i=1,...,n$ then $(x_i+V_{x_i})\times V\subseteq(x_i+V_{x_i})\times V_{x_i}$ too and so $$ x_i+V_{x_i}+ V=s[(x_i+V_{x_i})\times V]\subseteq s[(x_i+V_{x_i})\times V_{x_i}]=x_i+V_{x_i}+V_{x_i} $$ for each $i=1,...,n$ and so the last inclusion follows immediately. So could someone help me, please?
The remark after the Lemma shows that $0$ has a symmetric nbhd $V_x$ such that
$$V_x+V_x+V_x+V_x\subseteq(X\setminus C)-x\,,$$
since $(X\setminus C)-x$ is an open nbhd of $0$. And $0\in V_x$, so
$$0+V_x+V_x+V_x\subseteq(X\setminus C)-x\,,$$
and adding $x$ to both sides yields the desired result:
$$x+V_x+V_x+V_x\subseteq X\setminus C\,,$$
i.e.,
$$(x+V_x+V_x+V_x)\cap C=\varnothing\,.\tag{1}$$
Now suppose that $y\in(x+V_x+V_x)\cap(C+V_x)$; then there are $u,v,w\in C_x$ and $z\in C$ such that $x+u+v=z+w$, and hence $x+u+v-w=z$. But $V_x=-V_x$, so $-w\in V_x$, and therefore
$$x+u+v-w=z\in(x+V_x+V_x+V_x)\cap C\,,$$
contradicting $(1)$. It follows that
$$(x+V_x+V_x)\cap(C+V_x)=\varnothing\,.$$
I’m afraid that I can’t follow your subsequent argument; for starters, I don’t understand what $s$ is.