In proving a larger statement, I've got a claim:
Given that $C$ and $K$ are subsets of a topological vector space $X$ where $C$ is closed, $K$ is compact, and $x \in K\setminus C$, then $0$ has a symmetric neighbourhood $V_{x}$ such that $$ (x + V_{x} + V_{x} + V_{x}) \cap C = \emptyset. $$
I understand one proof of this (mostly): the proof starts by picking a point $x \in K$ and introduces a neighbourhood $W_{x}$ of $0$ where $K \subset x + W_{x}$ and $W_{x} \cap C = \emptyset$. A previous proposition is applied to introduce a symmetric neighbourhood $V_{x}$ of $0$ where $V_{x} + V_{x} + V_{x} + V_{x} \subset W_{x}$, from which the result follows.
I next want to prove that
The symmetry of $V_{x}$ shows then that $$ (x + V_{x} + V_{x}) \cap (C + V_{x}) = \emptyset, $$
but I'm not sure how to do this formally. I can see why it's the case pictorially, but the mathematical equivalent evades me at the moment.
Do it point-wise. If there is a point in the intersection, $y$ then we have $u,v, t\in V_x$ so that $x+u+v=y+t$, i.e. $x+u+v-t=y$. However $V_x+V_x+V_x+V_x\subseteq W_x$ and $t\in V_x\implies -t\in V_x$ by symmetry, hence this would give an element of $W_x$ which is also in $C$, a contradiction.