The isoperimetric problem in the plane

That is: among all closed curves in the plane of fixed perimeter, which has the largest area of enclosed region?

I have a sketchy proof that among smooth curves the answer is a circle. The proof probably resembles some existing ones, or may just be from my memory of a proof encountered almost 30 years ago.

My sketchy proof:

Let $C$ be an optimal curve and $A(C)$ be its enclosed area.

First of all, $C$ is non-concave, because otherwise a non-concave curve exists with the same perimeter and larger area.

Second, define diameter as a line segment that divides $A(C)$ by half. Clearly it also divides $C$ by half. Suppose a diameter $d$ intersects $C$ at points $x$ and $y$. Then the tangent lines to $C$ at $x$ and $y$ are both perpendicular to $d$, or else there is a concave optimal curve, a contradiction.

This implies that all diameters have the same length, and any two diameters intersect at their mid points, which always coincide. This implies $C$ is a circle.

Q.E.D.

The 3D version might be much more interesting!