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Well, as you have certainly already remarked (reading your post, I assume this), bilinearity makes a big difference. For the "only norm" case, your demonstration establishes that values on a dense set $U$ is sufficient (and then, we can reduce to countable). You cannot go further as, on a euclidean sphere $S_V$ a norm $p$ is the gauge function of the convex $$ C=\{x\in V| p(x)\leq 1\} $$ then, if your set $U$ is not dense, there is a point $M\in S_V$ and a neighbourhood $W$ of $M$ such that $U\cap W=\emptyset$. Now, deforming the sphere around $M$ in a convex way (you can find a one parameter deformation $C_t$ such), one obtains an infinite family of norms which coincides with $p$ on $U$ and differ from it.