WLOG, let the 6 vertices of the regular octahedron be $$A_1(0,0,1); A_2(0,0,-1); B_1(0,1,0); B_2(0,-1,0); C_1(1,0,0); C_2(-1,0,0).$$ Denote the regular octahedron as $Q$. Consider the point $P(x_0,y_0,z_0)$ where $x_0,y_0,z_0$ are nonnegative real numbers with $x_0+y_0+z_0>1$. Then $Q=|x|+|y|+|z|\le1$. (1) If $x_0+y_0\le1+z_0$, then there is a point $\left(\frac{x_0}{1+z_0},\frac{y_0}{1+z_0},0\right)$ on $PA_2$ in $Q$. Then we cannot see the edges $A_2B_1,A_2B_2,A_2C_1,A_2C_2.$ Therefore we can see maximum 8 edges from $P$. Similarly, when $y_0+z_0\le1+x_0$ or $z_0+x_0\le1+y_0$, we can also only see 8 edges from $P$. Now observe that what is left in the $1^{st}$ octant is $\{(x_0,y_0,z_0)|x_0+y_0>1+z_0, y_0+z_0>1+x_0, z_0+x_0>1+y_0\}$. For $P$ in this region, we can obviously see $A_1B_1, B_1C_1,C_1A_1$. Consider the plane $PA_1B_2$ whose equation is $$\frac{z_0-y_0-1}{x_0}\cdot x+y-z+1=0,$$ so $z=\frac{z_0-y_0-1}{x_0}\cdot x+y+1.$ (2) Combine (1) and (2), we obtain the following: $$1\ge\left|\frac{z_0-y_0-1}{x_0}\cdot x+y+1\right|+|x|+|y|\ge\left|\frac{z_0-y_0-1}{x_0}\cdot x+y+1+x-y\right|=\left|\frac{x_0+z_0-y_0-1}{x_0}\cdot x_0+1\right|.$$ Since $\frac{x_0+z_0-y_0-1}{x_0}>0, x\le0$. Also, since all points on the plane $PA_1B_2$ are nonnegative on the $x-axis$ and are $0$ on the $x-axis$ only if they are on the segement $A_1B_2$. By definition, $\triangle PA_1B_2\cap Q=\overline{A_1B_2}$, then we can see $A_1B_2$ from $P$. Similarly, we can also see $A_1C_2, B_1A_2,B_1C_2,C_1A_2,C_1B_2$ from $P$. Consider the plane $PA_2B_2$, whose equation is $-\frac{z_0+y_0+1}{x_0}\cdot x+y+z=0$. There is a point $\left(\frac{x_0}{1+y_0+z_0},0,0\right)$ on the plane inside $Q$. Therefore we cannot see $A_2B_2$ from $P$, and similarly, we also cannot see $B_2C_2, C_2A_2$ from $P$