Let $1>\epsilon > 0$ be given. Let $w\in V$ be arbitrary. Since you know that $C^\infty$ is dense in $V$, let $f \in C^\infty$ be a smooth function such that
$\Vert w - f \Vert_2 < \epsilon/2$
Now since $f$ is continuous on the compact set $[0,1]^d$, by uniform continuity there exists $\delta >0$ such that for all $x,y \in [0,1]^d$
$|x-y| < \delta \; \implies \; |f(x) - f(y)| < \epsilon/2$
Now let $h$ be small enough, so that on each cube of the grid, the maximal distance of points is smaller than $\delta$. Then by choosing $w^h \in V^h$ to take on an arbitrary value of $f$ in each cube where $w^h$ is constant, we have
$\Vert f - w^h \Vert_\infty^2 < \left(\epsilon/2\right)^2 \le \epsilon/2 $
Hence
\begin{eqnarray*} \Vert w - w^h \Vert_2 &\le& \Vert w - f\Vert_2 + \Vert f - w^h \Vert_2 \\ &\le& \Vert w - f\Vert_2 + \lambda([0,1]^d) \cdot \Vert f - w^h \Vert_\infty \\ &<& \frac \epsilon 2 + \frac \epsilon 2 = \epsilon \end{eqnarray*}
So that $V^h$ must be dense in $V$.