Understanding a proposition abount the number of regions a circle is divided into by $n$ points on its circumference My course notes (Mathematics BSc, third-year combinatorics module, unpublished) have a proposition,

Suppose that we have points $a_1,\dots,a_n$ in anticlockwise order around the unit circle, and we draw a line $l_{ij}$ from $a_i$ to $a_j$ for each $i\neq j$. Suppose that the points are in general position, so there is no point where more than two of the lines cross. Then the resulting diagram has $\binom{n}{2}$ lines, and $\binom{n}{4}$ interior crossing points, and $1+\binom{n}{2}+\binom{n}{4}$ regions.

The proof includes this diagram.

A couple of things that puzzle me here.

*

*The natural interpretation of "we draw a line $l_{ij}$ from $a_i$ to $a_j$ for each $i\neq j$ seems to me to be "draw every possible connecting line". None of the versions (which I'll call S2V1, S2V2, S2V3) of stage 2 show that, so I guess that's not the intended meaning. S2V2 seems to imply that different points can have different numbers of partners.

*S2V3 shows $C=0\neq1=\binom{4}{4}$, which seems to contradict the proposition.

Could someone clarify?
Edit 1: I'll add the section of the proof containing this diagram for clarity. The first part of the proof deals with lines and crossing points; the part I'm puzzling over deals with regions.

Edit 2: A case with an inner point (the centre) where more than two lines cross.

 A: The figure serves merely as an illustration of the first few steps of a process that would eventually add all the secants, one after the other. It is not a depiction of the final end state described in the statement. Many of the lines are still missing in the pictures for stage 2.
The parentheses below the diagrams seem to be quadruples of (number of lines $L$, number of inner points $C$, number of regions $R$, Euler characteristic of the disk $E$). The proof is establishing some more general relationship between these numbers, namely that $E=R-L-C=1$ will hold after each step of adding a single line, and no matter the order in which the lines are added.
Considering situations where not all the lines are included allows examining a process of adding one line at a time. This lends itself nicely to proof by induction. So by discussing a property for a more general class of arrangements, the proof becomes in fact easier.
The three images for stage 2 illustrate the three qualitatively different situations how the second line can lie with respect to the first. The two lines (when extended infinitely) can intersect within the circle, on the circle or outside the circle. In each of the cases, more lines would need to be added later on to achieve the situation from the statement. The different alternatives illustrate how the numbers of regions and crossings might differ for intermediate steps depending on the order in which you add lines, yet the combined formula $E$ stays the same in all five pictures.
In the final situation, then number of internal crossings is known and no longer depends on the order in which the lines were added. So by knowing $L=\binom n2$ and $C=\binom n4$ you can determine $R$ after showing that $E=1$ remains unchanged throughout.

we draw a line $l_{ij}$ from $a_i$ to $a_j$ for each $i\neq j$


[…] there is no point where more than two of the lines cross.

The second part refers to inner crossing points. Of course there are $n-1$ lines terminating at each of the $n$ points on the boundary, one for each other point. That's part of the setup, expressed in the first cited sentence. But each inner point is the intersection of exactly two lines.
For each pair of boundary points, you get a line connecting them. That's the $L=\binom n2$.
And for each combination of 4 boundary points, you get a total of 6 lines. But only two of them, the “diagonals”, intersect inside of the circle. So each set of four boundary points corresponds to exactly one inner point, thus $C=\binom n4$.
