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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, what you are looking for, if I understand correctly your question, is a set of uniqueness for the admissible norms on a given vector space $V$. Your demonstration establishes that values on a dense set $U$ is sufficient (and then, we can reduce to countable). You can go further by choosing a set $U$ such that $U\cup (-U)$ is dense, then you cannot go further as, on an euclidean sphere $S_V$, a norm $p$ is the jauge function of the balanced convex $$ C=\{x\in V|\, p(x)\leq 1\} $$ then, if your set $U\cup (-U)$ is not dense, there is a point $M\in S_V$ and a neighbourhood $W$ of $\{M,-M\}$ such that $U\cap W=\emptyset$. Now, deforming the sphere around $\{M,-M\}$ 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. If the space $V$ is complex, just replace the "real balanced saturation" $U\mapsto U\cup (-U)$ by its analogue (the orbit of $U$ under the group of complex numbers of modulus one $\mathbb{U}=\{z\in\mathbb{C} | |z|=1\}$).