Let $\nu$ be a norm on $V$ with unit $\nu$-ball $B_\nu = \{\textbf{x} \in V \mid \nu(\textbf{x}) \leq 1\}$. I want to show that the unit ball is closed using the definition. Let $\{x_n\} \subseteq B_\nu$ such that $x_n \to x$. We claim that $x \in B_\nu$. Since $\{x_n\}$ converges, for all $\epsilon > 0$, there exists $N$ such that for all $n \ge N$, $\nu(x - x_n) < \epsilon$. By the triangle inequality, $|\nu(x) - \nu(x_N)| < \epsilon \Rightarrow \nu(x_N) - \epsilon < \nu(x) < \epsilon + \nu(x_N)$. Since $\nu(x_N) \le 1$ and $\epsilon$ is arbitrary, $\nu(x) \le 1 \Rightarrow x \in B_\nu \Rightarrow B_\nu$ is closed.
Is my proof correct? I am not too sure about the part that reasons $\nu(x)\le 1$. Is there another way to make my proof more coherent? Thanks in advance!