Consider a circle inscribed in a square as shown below.

By comparing areas, it is easy to see that $\pi < 4$ from this figure.
What is the simplest proof using lengths to show that $\pi < 4$? Clearly, this is equivalent to showing that the length of the circular arc from $A$ to $B$ is less than the sum of the lengths of the straight line segments from $A$ to $C$ and from $C$ to $B$. I am interested in as elementary a proof as possible, either accessible to the Greek geometers or as close in spirit as possible to their methods.
Your idea can work, the fact that the square perimeter is larger than a circumference of an inscribed circle is so intuitive that it does not sound as a problem at the first glance.
The reason why it feels so obvious is following generalization: If P, Q are convex bodies and Q is a subset of P, then Q has smaller perimeter.
It can be proved inductively for polygons by triangle inequality and then you can generalize it by approximating the circle by polygons. But I think that Greek geometers would use another justification: Imagine Q as a solid object and P as a string. Then, if the string is shortened it get more tightly to Q up to its convex hull.