The following proof of your claim uses a somewhat different form of
"orthogonality", namely of translations of a certain function
on the Fourier side.
In the following, I use the version $\widehat{f}\left(\xi\right)=\int f\left(x\right)e^{-2\pi i\left\langle x,\xi\right\rangle }\, dx$
of the Fourier transform. It might be that I miss some factors of
$2\pi$ below. I hope that can be excused.
Let us fix $\varphi\in\mathcal{S}(\mathbb{R}^{d})$ with
the property that $\widehat{\varphi}\in C_{c}^{\infty}\left(B_{1}\left(0\right)\right)$
with $\widehat{\varphi}\geq0$ and $\varphi\left(0\right)=\int\widehat{\varphi}\, dx=2$.
Here, the first equality comes from Fourier inversion. Existence of
such a function is a standard result.
As $\varphi$ is continuous with $\varphi\left(0\right)=2$, there
is some $r>0$ such that $\left|\varphi\left(x\right)\right|\geq1$
holds for all $\left|x\right|\leq r$. By shrinking $r$, we can assume
$4\pi r\leq1$. Then, the family $\left(x_{k}\right)_{k}$ is a fortiori
also $\frac{4\pi r}{R}$-separated.
Consider $$\varphi_{R}\left(x\right):=\left(\frac{r}{R}\right)^{d}\cdot\varphi\left(\frac{r}{R}x\right).$$
We then have (easy calculation) $\left\Vert \varphi_{R}\right\Vert _{2}=\left(\frac{r}{R}\right)^{d/2}\cdot\left\Vert \varphi\right\Vert _{2}$,
as well as $\left|\varphi_{R}\left(x\right)\right|\geq\left(\frac{r}{R}\right)^{d}$
for $\left|x\right|\leq R$ and the Fourier transform of $\varphi_{R}$
is given by
$$
\widehat{\varphi_{R}}\left(\xi\right)=\widehat{\varphi}\left(\frac{R}{r}x\right)
$$
which is supported in $B_{r/R}\left(0\right)$.
For $\xi\in\mathbb{R}^{d}$ and $f:\mathbb{R}^{d}\rightarrow\mathbb{C}$,
write $\left(M_{\xi}f\right)\left(y\right):=e^{2\pi i\left\langle \xi,y\right\rangle }$
and $\left(L_{\xi}f\right)\left(y\right):=f\left(y-\xi\right)$. As
modulation corresponds to translation on the Fourier side (this is
an easy calculation), we have, for $k\in\left\{ 1,\dots,N\right\} $:
$$
\widehat{M_{\frac{x_{k}}{2\pi}}\varphi_{R}}=L_{\frac{x_{k}}{2\pi}}\widehat{\varphi_{R}}.
$$
As the family $\left(x_{k}\right)_{k}$ is $\frac{4\pi r}{R}$-separated,
we get
\begin{eqnarray*}
{\rm supp}\left(\widehat{M_{\frac{x_{k}}{2\pi}}\varphi_{R}}\right)\cap{\rm supp}\left(\widehat{M_{\frac{x_{n}}{2\pi}}\varphi_{R}}\right) & = & {\rm supp}\left(L_{\frac{x_{k}}{2\pi}}\widehat{\varphi_{R}}\right)\cap{\rm supp}\left(L_{\frac{x_{n}}{2\pi}}\widehat{\varphi_{R}}\right)\\
& \subset & \left[B_{r/R}\left(0\right)+\frac{x_{k}}{2\pi}\right]\cap\left[B_{r/R}\left(0\right)+\frac{x_{n}}{2\pi}\right]\\
& = & \frac{1}{2\pi}\cdot\left(\left[B_{2\pi r/R}\left(0\right)+x_{k}\right]\cap\left[B_{2\pi r/R}\left(0\right)+x_{n}\right]\right)=\emptyset.
\end{eqnarray*}
Finally, for $f_{1},\dots,f_{n}\in L^{2}(\mathbb{R}^{d})$ with
disjoint supports, we have
$$
\left\Vert \sum f_{j}\right\Vert _{2}^{2}=\int\left|\sum f_{j}\right|^{2}\, dx=\int\sum\left|f_{j}\right|^{2}\, dx=\sum\left\Vert f_{j}\right\Vert _{2}^{2}.\qquad\left(\dagger\right)
$$
We now employ $\left(\dagger\right)$, the Plancherel theorem, as
well as $\left|\varphi_{R}\left(x\right)\right|\geq\left(\frac{r}{R}\right)^{d}$
on $B_{R}\left(0\right)$, to derive
\begin{eqnarray*}
\left\Vert \sum_{k}a_{k}e^{i\left\langle x_{k},\xi\right\rangle }\right\Vert _{L^{2}\left(B_{R}\left(0\right)\right)} & \leq & \left(\frac{R}{r}\right)^{d}\cdot\left\Vert \sum_{k}a_{k}e^{i\left\langle x_{k},\xi\right\rangle }\varphi_{R}\left(\xi\right)\right\Vert _{L^{2}}\\
& = & \left(\frac{R}{r}\right)^{d}\cdot\left\Vert \sum_{k}a_{k}\cdot M_{\frac{x_{k}}{2\pi}}\varphi_{R}\right\Vert _{L^{2}}\\
& \overset{\text{Plancherel}}{=} & \left(\frac{R}{r}\right)^{d}\cdot\left\Vert \sum_{k}a_{k}\cdot\widehat{M_{\frac{x_{k}}{2\pi}}\varphi_{R}}\right\Vert _{L^{2}}\\
& \overset{\left(\dagger\right)}{=} & \left(\frac{R}{r}\right)^{d}\cdot\sqrt{\sum_{k}\left\Vert a_{k}\cdot\widehat{M_{\frac{x_{k}}{2\pi}}\varphi_{R}}\right\Vert _{L^{2}}^{2}}\\
& \overset{\text{Plancherel}}{=} & \left(\frac{R}{r}\right)^{d}\cdot\sqrt{\sum_{k}\left|a_{k}\right|^{2}\cdot\left\Vert \varphi_{R}\right\Vert _{2}^{2}}\\
& = & \left(\frac{R}{r}\right)^{d}\cdot\left(\frac{r}{R}\right)^{d/2}\cdot\left\Vert \varphi\right\Vert _{2}\cdot\sqrt{\sum_{k}\left|a_{k}\right|^{2}}\\
& = & \frac{\left\Vert \varphi\right\Vert _{2}}{r^{d/2}}\cdot R^{d/2}\cdot\sqrt{\sum_{k}\left|a_{k}\right|^{2}}.
\end{eqnarray*}
This proves the claim, as long as we allow the implied constant to
depend on the dimension $d$, as $\varphi,r$ can be chosen once and
for all for each dimension $d\in\mathbb{N}$. Also note that we did
not make use of the assumption $\left|x_{k}\right|=1$ for all $k$.
$$ |\sum a_ke^{i y_k \cdot \zeta}|{L^2(B(0,1))}\lesssim \left(\sum_k|a_k|^2\right)^{1/2} $$ So we have to bound the norm of the matrix with entries $c{k,l}=\int_{B(0,1)} e^{i (y_k-y_l) \cdot \zeta},d\zeta $ by a constant independent of $R$... – Jul 17 '14 at 04:54