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 $U$ (on the unit sphere) 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 $V$, 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 $M\in S_V$ (the unit sphere) 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 coincide 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 $\mathbb{U}=\{z\in\mathbb{C} | |z|=1\}$). 

> **Proposition** In order $U\subset S_V$ be a uniqueness set for the collection of all norms, it is necessary and sufficient that the orbit $G.U$ be dense in $S_V$ ($G=\{−1,1\}$ for the real case and $G=\mathbb{U}$ for the complex case).

[$G.U$ dense$\Longrightarrow$ $U$ is a uniqueness set] is clear. 

Now, if $G.U$ is not dense, it exists a point $w\in S_V$ which does not belong to $\overline{G.U}=G.\overline{U}$ and then 
$$
max_{v\in U}\,\{|\langle w|v\rangle|\}=M<1
$$
Now, the sets 
$$
C_t=\{v\in V|\, ||v||\,\leq 1\ \mathit{and}\ |\langle w|v\rangle|\leq (1-t)M+t\}
$$ 
for $0<t\leq 1$ gives the desired deformation (for each $C_t$ contains $U$).