Unit ball in finite dimension

If $B$ is the closed unit ball of $\R^{n}$ for some norm, then $B$ is a compact, symmetric ($i.e.$ stable by $x \mapsto -x$), convex neighborhood of $0$ in $\R^{n}.$ Those properties are independent of the norm, because they rely only on the vector space structure of $\R^n$ and its vector space topology.
Conversely, let $B \subseteq \R^n$ with the above properties. Using properties of the vector space topology on $\R^{n},$ for any nonzero $x\in \R^{n}$ the map
$$\gamma \colon \begin{cases} \R_{+} & \longrightarrow & \R^{n} \\ t & \longmapsto & tx \end{cases}$$is a homeomorphism; in particular, $\gamma^{-1}(B)$ contains a neighborhood of $0 \in \R^{+}.$ It also has convex image. Thus, the set $\{0\} \subsetneq I=\{ y \in B \ | \ \exists \lambda \in \R_{+} \colon y=\lambda x \}$ is a compact line segment, and we define $N(x) \in \R_{+}$ the only scalar such $I$ equals the segment $[0,\frac{1}{N(x)}x].$ We also set $N(0)=0$.
It is clear that $N(\lambda x) = \lambda N(x)$ for every positive scalar $\lambda,$ and one gets $N(-x)=N(x)$ from the fact that $B$ is symmetric. From this one can see that $B= \{ x \in \R^{n} \ | \ N(x) \leq 1\}.$ Finally, the triangle inequality stems from convexity of $B.$