Proof: Let there be $n$ planes parallel to the $Oxy$ plane containing points from $S$, denoted as $M_{1}, M_{2}, \cdots, M_{n}$. For the plane $M_{i}, 1 \leqslant i \leqslant n$, let it intersect the $Ozx, Ozy$ planes at lines $l_{y}$ and $l_{x}$, respectively, and let $M_{i}$ contain $m_{i}$ points from $S$. Clearly, $m_{i} \leqslant\left|S_{z}\right|$.

Let the sets of orthogonal projections of points on $M_{i}$ onto $l_{x}$ and $l_{y}$ be $A_{i}$ and $B_{i}$, respectively, and let $a_{i}=\left|A_{i}\right|$, $b_{i}=\left|B_{i}\right|$. Then, $m_{i} \leqslant a_{i} b_{i}$. Furthermore, since
$$\sum_{i=1}^{n} a_{i}=\left|S_{y}\right|, \sum_{i=1}^{n} b_{i}=\left|S_{x}\right|, \sum_{i=1}^{n} m_{i}=|S|,$$

by the Cauchy-Schwarz inequality, we have
$$\begin{aligned} \left|S_{x}\right| \cdot\left|S_{y}\right| \cdot\left|S_{z}\right| & =\left(\sum_{i=1}^{n} b_{i}\right)\left(\sum_{i=1}^{n} a_{i}\right) \cdot\left|S_{z}\right| \\ & \geqslant\left(\sum_{i=1}^{n} \sqrt{a_{i} b_{i}}\right)^{2} \cdot\left|S_{z}\right| \\ & =\left(\sum_{i=1}^{n} \sqrt{a_{i} b_{i}\left|S_{z}\right|}\right)^{2} \\ & \geqslant\left(\sum_{i=1}^{n} m_{i}\right)^{2}=|S|^{2} \end{aligned}$$

Thus, the proof is complete.