I'm currently reading through Convex Analysis and Beyond by Mordukhovich and Nam where the following preposition and proof are given (note that the authors define TVSs to be $T_0$).
Proposition 1.92 Let $X$ be a topological vector space $X$. Then $X$ is regular/$T_3$ while being, in particular, a Hausdorff topological space.
Proof. It suffices to show that for any closed set $F ⊂ X$ and $x ∉ F$, there exists an open neighborhood $V$ of the origin such that $(x + V) ∩ (F + V) = ∅$. Since $x ∈ F^c$ for the open complement $F^c$, we find a balanced open neighborhood $U$ of the origin with $x + U ⊂ F^c$. Then there exists a balanced open neighborhood $V$ of the origin such that $x + V + V ⊂ F^c$. It easily follows from the symmetry of $U$ that $(x + V) ∩ (F + V) = ∅$, which therefore verifies the statements of the proposition.
I'm having a hard time wrapping my head around the part in bold. What does the syymmetry of $U$ have to do with the two sums being disjoint?
Suppose $z \in (x+V)\cap (F+V)$. Then, $z=x+v=f+v'$ with $v,v' \in V, f \in F$. Now, $-v'\in V$, so $f=x+v+(-v')\in x+V+V \subset F^{c}$, a contradiction.