We wish to prove that
$W = \bigcap_{i=1}^{r} N_{g_{i}}$
Step $1$: Proving $W \subset \bigcap_{i=1}^{r} N_{g_{i}}$
Let $w \in W$. We know that the annihilator $W^{o}$ is the set of linear functionals that vanish on $W$. If $g_{i}$ is in the basis for $W^{o}$, it is certainly in $W^{o}$. Thus, the $g_{i}$ all vanish on $W$, so that $W \subset N_{g_i}$. We thus see that $w \in N_{g_{i}}$ for all $1 \leq i \leq r$. Hence
$w \in \bigcap_{i=1}^{r} N_{g_{i}}$
But $w$ was arbitrary, so
$W \subset \bigcap_{i=1}^{r} N_{g_{i}}$
Step $2$: Proving $W \supset \bigcap_{i=1}^{r} N_{g_{i}}$
Let $\{\alpha_{1}, \cdots, \alpha_{s}\}$ be a basis for $W$, and extend it to a basis for $V$, $\{ \alpha_{1} ,\cdots, \alpha_{n}\}$. Likewise, extend $\{g_{1}, \cdots, g_{r}\}$ to a basis for $V^{*}$, $\{g_{1}, \cdots, g_{n}\}$, noting that $\mbox{dim}(V) = \mbox{dim} (V^{*})$. It is easily seen that one can choose these two bases to be dual to each other, as the dual basis to $\{\alpha_{1}, \cdots, \alpha_{s}\}$ is not contained in $W^{o}$, and the dual basis for $V \setminus W$ must be contained in $W^{o}$, just by the nature of the dual basis.
(To make it easier to choose the correct basis for $V \setminus W$, try looking at the double dual $V^{**}$, which is naturally isomorphic to $V$, and looking at the dual basis for $W^{o}$ in $V^{**}$)
We get
$ g_{i}(\alpha_{j}) = \left\{ \begin{array}{cc} 1 & i= j\\ 0 & i \neq j \end{array} \right.$
Let $v \in \bigcap_{i=1}^{r} N_{g_{i}}$, so that $g_{i}(v) = 0$ for all $1 \leq i \leq r$, and write
$v = c_{1}\alpha_{1} + \cdots + \cdots c_{n}\alpha_{n}$
Taking $g_{i}$ of both sides, where $i$ ranges from $1$ through $r$,
$g_{i}(v) = c_{1}g_{i}(\alpha_{1}) + \cdots + c_{n}g_{i}(\alpha_{n}) = 0$
However, these $\{ g_{i}: 1 \leq i \leq r\}$ form a dual basis to $V \setminus W$. Hence, if $v$ had a nonzero component $c_{k}\alpha_{k}$ in $V \setminus W$, it would not vanish on $g_{k}$, $1 \leq k \leq r$. Then $g_{k}(r) \neq 0$, contrary to what we have shown. Thus $v$ must be contained in $W$.