Proof. Let $U \subset E$ be convex and balanced such that $U$ absorbs every bounded set of $E$. Let $\seq{U_n}\subset \cn^{o}(0)$ be a decreasing countable fundamental system of neighbourhoods at $0$. If $U_{n} \setminus nA \ne \emptyset$ for all $n \in \natp$, then there exists $\seq{x_n}$ such that $x_{n} \in U_{n} \setminus nA$ for all $n \in \natp$. In which case, $x_{n} \to 0$ as $n \to \infty$, so $\seq{x_n}$ is bounded. By assumption, there exists $n \in\natp$ such that $nA \supset \seq{x_n}$, which contradicts the fact that $\seq{x_n}\cap A = \emptyset$.$\square$