When we reflect a point $x$ on the number line in a point $y$, we get the reflected point $S_{y}(x)=y-(x-y)=2 y-x$. The same applies to points in space: if we reflect an arbitrary point $\left(x_{1}, x_{2}, x_{3}\right)$ in a point $\left(y_{1}, y_{2}, y_{3}\right)$, we get $S_{\left(y_{1}, y_{2}, y_{3}\right)}\left(x_{1}, x_{2}, x_{3}\right)=\left(2 y_{1}-x_{1}, 2 y_{2}-x_{2}, 2 y_{3}-x_{3}\right)$.
We now consider only points with integer coordinates. No matter which grid point ( $y_{1}, y_{2}, y_{3}$ ) we choose, the coordinates of the image $S_{\left(y_{1}, y_{2}, y_{3}\right)}\left(x_{1}, x_{2}, x_{3}\right)$ of $\left(x_{1}, x_{2}, x_{3}\right)$ all have the same parity as the coordinates of $\left(x_{1}, x_{2}, x_{3}\right)$. There are $2^{3}=8$ different possibilities for the parity combinations of the grid points. We can thus color the points of our grid with 8 colors, such that the color of a point is invariant under reflection in any grid point.
Consider now the cube with vertices $(0,0,0),(0,0,1),(0,1,0),(0,1,1),(1,0,0),(1,0,1)$, $(1,1,0),(1,1,1)$. These vertices each have a different color in the aforementioned coloring. Therefore, if one vertex is missing, we cannot obtain it by repeatedly reflecting one of the other vertices, because you only get points of a color you already had.