= Solution
Choose a <basis> $u_1,\ldots,u_m$ of $U$. Because $v\notin U$, the list $u_1,\ldots,u_m,v$ is <linearly independent>. Extend it to a <basis> of $V$. A <linear functional> is specified uniquely by its values on a <basis>, so define $f$ to be zero on the $u_j$ and on every added basis vector other than $v$, and set $f(v)=1$.
Then $f$ vanishes on every <linear combination> of the $u_j$, so $f\in U^0$, while
$$
\boxed{f(v)=1\neq0.}
$$
This proves that the <annihilator of a vector subspace> separates every vector outside the subspace from that subspace; no <inner product> or preferred coordinates are needed.
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