Can all polygons outside of the largest inscribed rectangle in a convex polygon be concave

Explanation:
C is the hexagon with vertices $(0,\pm <!--\; \pm \; -->1)$, $(\pm <!--\; \pm \; -->(1+\mathrm{ϵ<!--\; ϵ\; -->}),\pm <!--\; \pm \; -->\mathrm{ϵ<!--\; ϵ\; -->})$, where $\mathrm{ϵ<!--\; ϵ\; -->}\ll <!--\; \ll \; -->1$. I conjecture that the largest rectangle inscribed in C is the square with vertices $(\pm <!--\; \pm \; -->1,0)$, $(0,\pm <!--\; \pm \; -->1)$.

Step 1
Suppose there is such a shape C.
Consider the largest rectangle R in C. If 2 consecutive vertices of R touch C, then the part that is cut out will be convex. Hence, in order to fulfill your conditions, at least 1 of any 2 adjacent vertices must not lie on the boundary of C. This implies that 2 opposite vertices of R do not lie on the boundary of C.
If 4 vertices of R do not lie on the boundary of C, it is easy to see that a small expansion will result in a larger R′, contradicting the maximality of R.
If 3 vertices of R do not lie on the boundary of C, notice that due to convexity of C, the boundary of C can only touch the perimeter of R at that one vertex. Then, once again, a small expansion from that vertex will result in a larger R′, contradicting the maximality of R.
Hence, we must have 2 opposite vertices of R (labelled a,c) which do not lie no the boundary of C, and the other 2 (labeled b,d) will lie on the boundary of C.
Step 2
If the boundary of C touches ba or da at a point that is not b or d, then the convexity of C implies that the boundary of C touches a, which is a contradiction. As such, the boundary of C only touches R at the points b and d.
Let O be the center of R (intersection of diagonals). Consider the circle with center O and diameter bd. If ac is not the perpendicular (diameter) to bd, then by rotating points a,c with respect to O, we can find a rectangle bedf inscribed in O with a larger area.
This leaves us with abcd being a square. WLOG, let's use a coordinate system with $O=(0,0),b=(-<!--\; -\; -->1,0),a=(0,1),d=(1,0),c=(0,-<!--\; -\; -->1)$
Step 3
Since a and c do not touch the boundary of C, let $2\mathrm{ϵ<!--\; ϵ\; -->}$ be the minimum of the distance between either of points a or c with the boundary of C. Furthermore, set $\mathrm{ϵ<!--\; ϵ\; -->}<0.1$. Let $a^{\prime }=(0,1+\mathrm{ϵ<!--\; ϵ\; -->})$, $c^{\prime }=(0,-<!--\; -\; -->1-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->})$, neither of which points touch the boundary of C. By the convexity of C, the quadrilateral a′bc′d only touches C at b and d.
I feel that we should be able to find a rectangle of area $>2$ here, but am unable to do so easily.