However, I finally rewrote it for myself so that I could make sense of it. It's not perfect, but I figured I'd share what I wrote in case it's helpful:

Lemma 2
Consider all the arcs with a given length and whose endpoints are called S and T. The curve that encloses the maximum area between it and the straight line between S and T is a semicircle.

Proof of Lemma 2
The proof relies on two results (whose proofs are left as an exercise for the reader):
(a) For every arc with endpoints S and T, if all angles formed on a point of the arc by lines from that point through S and T are right angles, then the arc is a semicircle.
(b) Among all triangles with two given sides, the one where those two sides form a right angle has the largest area.

Now consider an arc with endpoints S and T for which there exists a triangle whose vertices are S, T, and a point on the arc, where the angle formed at that point is not a right angle (Figure 4). For convenience, we will call such an arc an "arc with a non-right angle." In that case, we could slide S (either to the left or right, depending on the whether the angle is acute or obtuse), so as to create a triangle where the angle is right. Let pieces of the arc move along (Figure 5). By result (b) above, the area of the triangle grew. Since the area of the two red regions did not change but the area of the triangle grew, the whole area between the new arc and the line ST has increased. Therefore, for any arc with a non-right angle, another arc of the same length can be formed enclosing a larger area between it and the line connecting its endpoints. Hence, all such arcs enclose less area than ones where the angle is always right . But by result (a), any arc where the angle is always right is a semi-circle.

By the way, I wonder if there's a way, instead of doing lemma 2, to just skip it altogether and extend lemma 1? Instead of dividing the closed curve into two, divide it into N segments, each with the same perimeter length, and two line segments going to the centroid. (e.g., like cutting out pieces of a pie). Then one could use similar logic as lemma 1 and show that as N approaches infinity, the pie pieces approach line segments, and if they are all equal, then one has a circle by definition.

I realize though that this involves the added complexity of proving that if the pie pieces are cut to the centroid, the reasoning paralleling in lemma 1 still holds (obviously it wouldn't if one just picked an arbitrary point).

I do not believe that you can split the shape into N pieces as you suggest - at least not easily. The requirement of them having the same perimeter is rather a stringent constraint, even if you choose as the nexus of the construction the centroid of the shape.