Exercise 2.6.19

Let $S^{\prime }$ is a basis for $W$ and we can extend it to be a basis $S$ for $V$. Since $W$ is a proper subspace of $V$, we have at least one element $t\in S$ such that $t\notin W$. And we can define a function $g$ in $F(S,𝔽)$ by $g(t)=1$ and $g(s)=0$ for all $s\in S$. By the previous exercise we know there is one unique linear functional $f\in V^{\text{∗}}$ such that $f_S=g$. Finally since $f(s)=0$ for all $s\in S^{\prime }$ we have $f(s)=0$ for all $s\in W$ but $f(t)=1$. So $f$ is the desired functional.