[proofplan] The proof is a direct verification of set inclusion. We take an arbitrary linear functional in $W^0$, unfold the definition of annihilator, and use the hypothesis $U \subset W$ to show that the same functional vanishes on every element of $U$. This proves membership in $U^0$, and since the functional was arbitrary, the desired inclusion follows. [/proofplan] [step:Take an arbitrary functional vanishing on $W$] Let $\lambda \in W^0$ be arbitrary. By the definition of $W^0$, the symbol $\lambda$ denotes a $k$-linear map \begin{align*} \lambda: V \to k \end{align*} such that $\lambda(w)=0$ for every $w \in W$. [/step] [step:Restrict the vanishing condition from $W$ to $U$] Let $u \in U$ be arbitrary. Since $U \subset W$, we have $u \in W$. Therefore the defining property of $\lambda \in W^0$ gives \begin{align*} \lambda(u)=0. \end{align*} Because $u \in U$ was arbitrary, $\lambda$ vanishes on every element of $U$. Hence, by the definition of $U^0$, we have $\lambda \in U^0$. [guided] We need to prove the inclusion $W^0 \subset U^0$. For a set inclusion, the correct method is to begin with an arbitrary element of the left-hand side and prove that it belongs to the right-hand side. Let $\lambda \in W^0$ be arbitrary. By definition of the annihilator of $W$ in $V^*$, this means that $\lambda$ is a $k$-linear map \begin{align*} \lambda: V \to k \end{align*} and that $\lambda$ vanishes on every vector of $W$: \begin{align*} \lambda(w)=0 \text{ for every } w \in W. \end{align*} To prove that $\lambda \in U^0$, we must prove that $\lambda$ vanishes on every vector of $U$. Let $u \in U$ be arbitrary. The hypothesis $U \subset W$ says precisely that every element of $U$ is also an element of $W$, so this particular vector $u$ lies in $W$. Applying the vanishing property of $\lambda$ on $W$ to this vector gives \begin{align*} \lambda(u)=0. \end{align*} Since the choice of $u \in U$ was arbitrary, $\lambda$ vanishes on all of $U$. Therefore $\lambda \in U^0$ by the definition of the annihilator of $U$. [/guided] [/step] [step:Conclude the inclusion] We have shown that every arbitrary element $\lambda \in W^0$ also belongs to $U^0$. Therefore \begin{align*} W^0 \subset U^0. \end{align*} This proves the theorem. [/step]