Proof. Firstly, (TVS2) implies that every neighbourhood of $0$ is circled.

By Proposition 5.1.16, $E$ admits a fundamental system of neighbourhoods consisting of open sets or closed sets.

Let $U \in \cn^{o}(0)$ be open. By (TVS2), there exists $r > 0$ such that $\lambda U \subset U$ for all $\lambda \in K$ with $\abs{\lambda}\le r$. Define

then for any $x \in V$, there exists $\lambda \in K$ with $\abs{\lambda}\le r$ and $y \in U$ such that $x = \lambda y$. In which case, for any $\mu \in K$ with $\abs{\mu}\le 1$, $\mu(\lambda y) = (\mu \lambda) y \in \mu\lambda U$. Since $\abs{\mu \lambda}\le r$, $\mu \lambda U \subset V$. Thus $V \subset U$ is balanced.

Let $U \in \cn(0)$ be closed, then there exists a balanced neighbourhood $V \in \cn^{o}(0)$ such that $V \subset U$. In which case, for any $\lambda \in K$ with $0 < \abs{\lambda}\le 1$, $\lambda \overline{V}= \overline{\lambda V}\subset \overline{V}$ by (TVS2). Therefore $\overline{V}\subset U$ is balanced as well.$\square$