jmc

$△ABC$ is a right isosceles ($45^{\circ }-45^{\circ }-90^{\circ }$) triangle, so $AC=AB\sqrt{2}=\sqrt{2}$. Thus, the side length of the octagon is $\sqrt{2}$.

We can compute the octagon's area by subtracting the area of the four right isosceles triangles from the area of square $BDEF$.

The four right isosceles triangles are congruent by symmetry and each has an area of $\frac{1}{2}\cdot 1\cdot 1$, so their total area is $$4\cdot \frac{1}{2}\cdot 1\cdot 1=2.$$ Each side of square $BDEF$ is comprised of a leg of a right isosceles triangle, a side of the octagon, and another leg of a different right isosceles triangle. Hence, the side length of $BDEF$ is $1+\sqrt{2}+1=2+\sqrt{2}$, and the area of $BDEF$ is $$(2+\sqrt{2}{)}^2=4+2+4\sqrt{2}.$$ Finally, the area of the octagon is $$4+2+4\sqrt{2}-2=\boxed{4+4\sqrt{2}}.$$