You have to be a bit more precise, there are several norms that induce the product topology on $V \times V$.
We have $$\|(v_1,v_2)\|_2 = \sqrt{\|v_1\|^2 + \|v_2\|^2}$$ as what you probably mean by the Euclidean one and $$\|(v_1, v_2)\|_1 = \|v_1\| + \|v_2\|$$
and $$\|(v_1,v_2)\|_\infty = \max\left(\|v_1\|,\|v_2\|\right)$$
as two other ones. All of these can checked to be valid norms on $V \times V$.
As we have $$\|(v_1,v_2)\|_\infty \le \|(v_1,v_2)\|_1 \le \|(v_1,v_2)\|_2 \le \sqrt{2}\|(v_1,v_2)\|_\infty$$ from the usual inequalities of reals, we can argue that all these norms on $V \times V$ are equivalent.
And as $B_{\infty}((v_1,v_2), r) = B(v_1,r) \times B(v_2,r)$ it's also not too hard to argue that the $\|\cdot\|_\infty$ surely induces the product topology. (open balls form a base, and products of open balls (from $V$) thus form a base for $V \times V$ etc.).
Best Answer
The definition of a norm already assumes the set is a vector space over $\mathbb{R}$ or $\mathbb{C}$. If we don't have addition and scalar multiplication then the axioms $||x+y||\leq ||x||+||y||$ and $||\alpha\cdot x||=|\alpha|\cdot ||x||$ make no sense.
You should not confuse between normed spaces and metric spaces. These are two different terms. The relation is that a norm defines a metric on the vector space by $d(x,y)=||x-y||$, but that's it. There are lots of other metrics as well. (and not just on vector spaces)