= Solution
Fix an integer $k>n$ and partition $[0,1]$, up to endpoints of measure zero, into
$$
I_p=[(p-1)/k,p/k),
\qquad 1\le p\le k.
$$
For $p\ne q$, define
$$
f_{p,q}=\sqrt{\frac{k}{2}}\,
(\mathbf1_{I_p}-\mathbf1_{I_q}).
$$
These functions have integral zero and $L^2$ norm one, so they are admissible.
Let
$$
N_p=\sum_{i=1}^n\mathbf1_{\{Z_i\in I_p\}}
$$
be the interval occupancy counts. Since each $Z_i$ is uniform, $\mathbb EN_p=n/k$, and hence
$$
D(f_{p,q})
=\frac{k}{2n^2}\mathbb E(N_p-N_q)^2.
$$
For every outcome,
$$
\begin{aligned}
\sum_{p,q=1}^k(N_p-N_q)^2
&=2k\sum_{p=1}^kN_p^2-2\left(\sum_{p=1}^kN_p\right)^2\\
&=2k\sum_{p=1}^kN_p^2-2n^2\\
&\ge2kn-2n^2,
\end{aligned}
$$
because the nonnegative integer counts satisfy $N_p^2\ge N_p$ and $\sum_pN_p=n$. No independence assumption was used.
Averaging over all ordered pairs gives
$$
\frac1{k^2}\sum_{p,q=1}^kD(f_{p,q})
\ge\frac1n-\frac1k.
$$
The diagonal terms vanish, so some admissible pair $p\ne q$ satisfies the same lower bound. Therefore
$$
\sup_{\substack{\|f\|_2=1\\\int f\,d\lambda=0}}D(f)
\ge\frac1n-\frac1k.
$$
Letting $k\to\infty$ proves the <dependent uniform sampling variance lower bound>
$$
\boxed{\sup D(f)\ge\frac1n}.
$$
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