= Solution
The unheaded preliminary requests are addressed before the first example. A <norm> is a function $p:V\to[0,\infty)$ satisfying $p(x)=0$ exactly for $x=0$, $p(ax)=|a|p(x)$, and $p(x+y)\le p(x)+p(y)$. Its induced metric is $d(x,y)=p(x-y)$. Translation $T_v$ preserves this distance, and scalar multiplication satisfies $d(ax,ay)=|a|d(x,y)$, so both are continuous; multiplication by zero is constant.
The <open unit ball> determines the norm uniquely through its <Minkowski functional>:
$$
p(x)=\inf\{t>0:x/t\in B\},
$$
because $x/t\in B$ is equivalent to $p(x)<t$. This also works for $x=0$.
Under the given radial and convexity hypotheses, set $p_B(0)=0$ and $p_B(v)=1/\lambda(v)$ for $v\ne0$. The finite symmetric segment along each line gives positivity, absolute homogeneity and $v\in B\iff p_B(v)<1$. If $s>p_B(x)$ and $t>p_B(y)$, then $x/s,y/t\in B$, and the stated convexity condition yields $(x+y)/(s+t)\in B$. Hence $p_B(x+y)<s+t$; taking $s\downarrow p_B(x)$ and $t\downarrow p_B(y)$ proves the <triangle inequality>. Thus this is a <norm determined by a convex radial unit ball>. The case $V=\{0\}$ is immediate.
For the coordinate box, the <Minkowski functional> is the <supremum norm>:
$$
\boxed{\|(x_1,\ldots,x_n)\|_B=\max_i|x_i|.}
$$
Indeed $x/t\in B$ exactly when $t>\max_i|x_i|$.
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