Let $\mathbb{T}$ denote the periodic interval $[-\pi, \pi)$.
Is the system of all (periodic) shifts of $\mathbb{I}_{[-h,h]}$, for some $h < \pi$, a complete system in $L_2(\mathbb{T})$?
Solution attempt: since the operator that takes a function $f(x)$ in $L_2(\mathbb{T})$ and sends it to the series of its complex Fourier coefficients $c_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{inx}dx$ is an isometry between $L_2(\mathbb{T})$ and $l^2(\mathbb{Z})$, a system is complete iff its image under this operator is also complete.
Taking $f(x)=\mathbb{I}_{[-h,h]}$, the coefficients of the shifts are $$f(x-a) \rightarrow c_{n,a}=\frac{\sin(nh)}{2\pi n}e^{ina}$$
Now what remains to show is that the system of series $\frac{\sin(nh)}{2\pi n}e^{ina}$, $a \in [-\pi, \pi]$, is complete in $l^2(\mathbb{Z})$. How to show this?
Suppose $f$ is $2\pi$ periodic and is orthogonal to all periodic translates of $\mathbb{I}_{[-h,h]}$. Then $$ 0=\int f\mathbb{I}_{[-h+x,h+x]}dt = \int_{-h+x}^{h+x}fdt,\;\; \forall x. $$ Setting $x=h$ gives $\int_{0}^{2h}fdx$ for all $h > 0$, which forces $f$ to be constant a.e.. And, that constant must be $0$, which gives $f=0$ a.e. by the Lebesgue differentiation theorem. So the closed linear span of all translates of $\mathbb{I}_{[-h,h]}$ for all $h$ must generate a dense linear subspace of $L^2$.