Largest-area shape with diameter 1?

I think it is the circle by the following reasoning. I'm not totally sure about this though, so if there is an error hopefully someone can correct me.

The shape of maximal area must be bounded by a curve of constant width (with the diameter being the width), since any curve of varying width could be expanded in the directions where the width is deficient, increasing the area while keeping the diameter the same.

Now by Barbier's theorem, the perimeter of any curve of constant width $w$ is just $\pi w$, regardless of the shape. Thus all shapes that can be candidates for optimality must have the same perimeter.

But by the isoperimetric inequality, a circle is the curve enclosing maximal area among all shapes with a given perimeter. Thus the circle encloses the maximal area among all shapes with a given diameter.

The isodiametric inequality states $V(C) \le \left(\frac{\operatorname{diam}(C)}{2}\right)^n V(B_2^n)$ for any bounded set $C \subset \mathbb{R}^n$. This means that a ball of radius $\frac{1}{2}$ has the biggest volume of all shapes with diameter at most $1$.

A prove can be found in Gruber's book about convex geometry. Wlog you can assume $C$ to be convex and compact. It boils down to applying successive Steiner-symmetrizations with respect to the coordinate hyperplanes to $C$. The result is a convex body $D$ that has the same volume as $C$, the diameter of $D$ is at most the diameter of $C$ and $D$ is symmetric with respect to the origin. The last part implies that $D$ is a subset of $\frac{\operatorname{diam}(D)}{2} B_2^n$.