This is not a full proof - it is a summary of my ideas thus far. Let $X$ be the area of the octagon and $T$ be the total area of the $8$ triangles around the octagon. $2X=1+1-T \implies X=1-\frac{T}{2}$ so we have to show that $T \le 1$. Fold in every second triangle around the octagon. Consider two of the adjacent folded in triangles. We can easily angle chase and establish some parallel results, from which it follows that these triangles do not intersect. For the opposite triangles that have been folded in, I don't yet have a proof for why they don't intersect, but it appears intuitively obvious to me. However, that clearly doesn't constitute a proof. Once we have proven that none of the folded in triangles intersect each other, all of the $8$ triangles around the octagon lie, pairwise non-intersecting, in one of the rectangles, which we know has area $1$. Thus $T \le 1$, and we are done.

I think I have a complete solution to this very nice problem. My idea is similar to ACCCGS8 one, but not inspired by his post. Let $A_1A_2A_3A_0$ and $B_1B_2B_3B_0$ be the given rectangles. Denote by $s(X)$ the area of a figure $X$ and the indices will be taken modulo $4$. Take $\{C_i\}=A_iA_{i-1}\cap B_iB_{i-1}$ and $\{D_i\}=A_iA_{i+1}\cap B_iB_{i-1}$ for all $i=0,1,2,3$. Moreover, when I write $a_i$ I will refer to $\triangle A_iC_iD_i$ and the same way $b_i$ means $\triangle B_iC_{i+1}D_i$. Also, $q$ will be our octagon. Define now $X_i$ and $Y_i$ such that $A_iC_iX_iD_i$ and $B_iC_{i+1}D_iY_i$ are rectangles. For triangles $\triangle X_iC_iD_i$ and $\triangle Y_iC_{i+1}D_i$, the meaning of $x_i$ and $y_i$ will be similar to $a_i$ and $b_i$. We have $s(a_1)+s(a_2)+s(a_3)+s(a_0)=s(b_1)+s(b_2)+s(b_3)+s(b_0)=1-s(q)$, hence $s(x_1)+s(x_2)+s(x_3)+s(x_0)=s(y_1)+s(y_2)+s(y_3)+s(y_0)=1-s(q)$. We can easily see that triangles $x_i$ and $x_{i+1}$ can not overlap since they are separated by a band perpendicular to $D_iC_{i+1}$. The same argument works for triangles $y_i$ and $y_{i+1}$. We can also see that triangles $x_i$ and $y_i$ can not overlap since the angle $\widehat{A_iD_iB_i}$ is obtuse and the same argument works for triangles $x_i$ and $y_{i-1}$. Now if triangles $x_i$ and $x_{i+2}$ do not overlap for $i=0,1$, then we are done because those triangles are situated inside $q$ and then $2s(q)>2s(q)+s(x_1)+s(x_2)+s(x_3)+s(x_0)=1$. So suppose that $x_j$ and $x_{j+2}$ overlap for some $j$, then $y_i$ and $y_{i+2}$ can not overlap since there will be a band made by the sides of triangles $x_j$ and $x_{j+2}$ that will separate them. Therefore we get $2s(q)>2s(q)+s(y_1)+s(y_2)+s(y_3)+s(y_0)=1$ and we are done.