Is the following Proof Correct?
Theorem. Given that $V$ is finite-dimensional and $U\subset V$. $U = \{0\}$ if and only if $U^0 = V'$
Proof. $(\Rightarrow).$ Assume that $U = \{0\}$ and $\phi\in V'$, evidently $\phi(0) = 0$ consequently $\phi\in U^0$ and since our choice of $\phi$ was arbitrary it follows that $U^0 = V'$.
$(\Leftarrow).$ For the converse assume that $U^0 = V'$ and $u\in U$ further more let $v_1,v_2,...,v_n$ be a basis of $V$ and let $\phi_1,\phi_2,...,\phi_m$ be the corresponding dual basis. Expressing $u$ as a linear combination of $v_1,v_2,...,v_n$ of $V$ yields the following equation. $$u = c_1v_1+c_2v_2+\cdot\cdot\cdot+c_mv_m$$ From hypothesis we know that $U^0 = V'$ then in particular $\phi_1\in U^0,\phi_2\in U^0,...,\phi_n\in U^0$ which implies $\phi_1(u) = c_1 = 0,\phi_2(u) = c_2 = 0,...,\phi_n(u) = c_n = 0$ consequently $u = 0$. Since our choice of $U$ was arbitrary it follows that $U = \{0\}$.
$\blacksquare$