How is the area of a circle calculated using basic mathematics?

I like this (entirely non-rigorous) approach:

(The image has been taken from this article on Wikipedia)

It should be "intuitively obvious" that increasing the number of sectors makes the approximation better and better.

Well, let's take a close look at what you're doing. Right now, if you want to calculate the area of a circle of radius $4$, you're summing up the circumferences in the following picture, times the "thickness" of each skin, which is $1$:
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However, this method has a problem: Each circumference represents the area of everything inside it, but before the next circle. So, for instance, the circumference of the red circle is approximating the area in which the blue circle lies in the following image:
$\hskip1.1in$
which is troublesome because the red circumference is bigger than the blue circumference - which is a problem, meaning your method is an overestimate.

More rigorously speaking, we need that the area of a circle is invariant of how we slice it - if we divide a circle into rings of half the thickness as follows:
$\hskip1.1in$
then your method suggest less area - in particular, if the original area your method obtained was $\pi (n^2+n)$, then the new area is $\pi (n^2+\frac{n}2)$ (A quick way to get this result without summing a new series is to see that this is the same as doubling $n$, and taking a quarter of the area thus calculated). This should trouble us some, because the area of a circle shouldn't depend on how we measure it - and it should scale so that doubling the radius quadruples the area. However, if we subdivided each of the original rings into $c$ sections, we'd get an area of $\pi (n^2+\frac{n}c)$ and, as we increase $c$, we get closer to the correct area of $\pi n^2$. In a rigorous sense, we may consider the process of dividing finer and finer sections of the circles to be an example of integration - but we can avoid such such tools by refining our geometric approach.

In particular, instead of estimating area of a slice as the outer ring times its thickness, we can estimate the area of a slice by a circle through the "midpoint" of the ring times the thickness - that is, we use the dotted lines in the following diagram to estimate the area of the sections through which they pass:
$\hskip1.1in$
which yields the sum: $$2\pi \cdot \frac{1}2 + 2\pi \cdot \frac{3}2 +2\pi \cdot \frac{5}2 +\ldots + 2\pi \cdot \frac{2n-1}2 $$ which equals $$\pi n^2.$$ This happens to be the correct answer - and we can convince ourselves of this through various methods. The most elementary is to consider that if we make finer and finer* divisions as before, this value does not change - hence, it must be the correct value as any other value would be shifted towards the actual value as we do this. Calculus also tells us that since the circumference of a circle increases linearly with its radius and we are integrating over this, the amount we over/under estimate by choosing the center balances out. In any case, this yields the correct answer and gives a plausible geometric reason for it to be so.

*We can also make more coarse divisions - we could have one division and guess $2\pi \cdot \frac{n}2 \cdot n$ as the circumference of a circle of half the radius times the "thickness" of the whole circle.

The problem is that your onion doesn't have enough layers. If you were to integrate the circumference function $2\pi r$, which is equivalent to making infinitely many layers, you would indeed get $$2\pi \cdot \frac{1}{2}r^2=\pi r^2$$ However, I'm not sure if you call integration "basic math".

Alternatively, you can see a circle as a regular $n$-gon with $n \rightarrow \infty$ sides and side length $k$. This polygon is approximated by its circumscribed circle with radius $r$. Of course, we can divide the polygon into $n$ triangles. We can approximate $k$ by $\frac{2\pi r}{n}$. The height is $r$. Therefore the total area of such a triangle is $\frac{1}{2}\frac{2\pi r}{n}r=\frac{\pi r^2}{n}$ therefore the total area of the polygon and thus the circle is $\pi r^2$. Of course, to write it down rigorously, we also have to consider the inscribed circle and write down bounds etcetera, but this is the idea.

You need to add areas, not lengths. So you approximate the area of the layers by multiplying the circumference times a small radial lenght: $$ \left (2\pi\,\frac rn\right)\,\frac rn,\ \ \left (2\pi\,\frac {2r}n\right)\,\frac rn,\ \ \ldots \ \ ,\left (2\pi\,\frac {nr}n\right)\,\frac rn. $$ The sum is $$ \frac {2\pi r^2}{n^2}\, (1+2+\cdots+n)=\pi r^2\,\left (1+\frac1n\right). $$