problem on annihilators on finite dimensional spaces 
Suppose $V$ and $W$ are subspaces of a finite-dimensional vector space $U$.
Show that if $V^0 \subset W^0$ then  $W \subset V$

This is an exercise problem in Linear Algebra Done Right, 3rd Edition, in Chapter 3.F . For a vector space $V$ over $K$, with $S$ being a subset, the annihilator of $S$ is the subspace $S^0$ of linear functionals $f$ in $V^*$ so that $f(s)=0$ for every $s$ in $S$. 
I would appreciate any hints. Note that at this point, more complicated ideas of double dual etc have not been introduced. This question has been asked as a sub-part of a problem earlier as Subspaces and annihilators but  this part has not been addressed.

Suppose that $W \nsubseteq V$, i.e. that there exists some $w \in W$ with $w \notin V$. Then clearly $w \neq 0$. So if we start with a basis of $(v_1, \dotsc, v_n)$ of $V$ the familiy $(v_1, \dotsc, v_n, w)$ is still linearly independent. We extend this to a basis $(v_1, \dotsc, v_n, w, u_{n+2}, \dotsc, u_m)$ of $U$. Now define $f \in U^*$ by $f(v_i) = 0$ for every $1 \leq i \leq n$, $f(w) = 1$ and $f(u_i) = 0$ for every $n+2 \leq i \leq m$.
Then $f(v) = 0$ for every $v \in V$ because $(v_1, \dotsc, v_n)$ is a basis of $V$ with $f(v_i) = 0$ for every $1 \leq i \leq n$. Therefore $f \in V^0$. But $f(w) = 1 \neq 0$ with $w \in W$, so $f \notin W^0$. 
