Prove that a $TVS$ $E$ is compact Hausdorff space if and only if it consists of a single element, $0$.
My attempt: The implication $(\Leftarrow)$ is trivial.
As for another implication, suppose that $E \neq \{0\}$ then there exists $y \in E$ such that $y \neq 0$. I was trying to build a balanced neighborhood of $y$ to arrive at a contradiction of the type $y \in \frac{1}{n}V$ for all $n \in \mathbb{N}$.
Since you're considering complex vector spaces (it would work the same for real, or even rational vector spaces, but for exotic scalar fields the argument need not work), we can complete your idea.
Suppose $y \in E \setminus \{0\}$. Let $V$ be an open balanced neighbourhood of $0$ that doesn't contain $y$. Such a $V$ exists since $E$ is assumed to be Hausdorff. Then $\mathscr{V} = \{nV : n \in \mathbb{N}\}$ is an open cover of $E$ that has no finite subcover.
Every finite subfamily $\mathscr{F}$ of $\mathscr{V}$ is contained in $n_1\cdot V$ for some $n_1$, but then $n_1\cdot y$ does not lie in any member of $\mathscr{F}$, thus $\mathscr{F}$ doesn't cover $E$.