To summarise, the flaw is that you're treating infinity as if it's something finite. You can't throw around infinities and infinitesimals too casually. My answer will have a few hand-wavy arguments, but it should be enough for you to realise why your argument was flawed.
In your argument you're taking the limit as side-lengths ($l$) go to zero and the limit as the number of sides ($n$) tend to infinity simultaneously. The rule for "simultaneous limits" is that there must be no dependence on order. This means you want:
\begin{align*}
\lim_{n\rightarrow\infty}\lim_{l\rightarrow0}P(n, l) = \lim_{l\rightarrow 0}\lim_{n\rightarrow\infty}P(n, l) = \mathrm{Circle}
\end{align*}
Where $P(n, l)$ is a regular polygon with side-lengths $l$, and $n$ sides.
Imagine you have an infinite number of unit-length line segments. If you use $3$ of them, you have an equilateral triangle with side lengths of $1$. If you use $n$ of them, you get the regular $n$ sided polygon with side lengths of $1$. Suppose you were to measure the interior angle of such a polygon. If this angle is less than $180^{\circ}$, then you can work out how many sides were used to construct it. In other words, it's still only a finite number of sides. Thus $\lim_{n\rightarrow\infty}P(n, l)$ must be equal to an infinite straight line. This no longer has any dependence on $l$, because $\infty/l = \infty$ for any $l\neq 0$. Therefore,
\begin{align*}
\lim_{l\rightarrow 0}\lim_{n\rightarrow\infty}P(n, l) = \lim_{l\rightarrow 0}(\lim_{n\rightarrow\infty}P(n, l)) = \lim_{l\rightarrow 0}(\mathrm{Infinite\ straight\ line}) = \mathrm{Infinite\ straight\ line}
\end{align*}
How about the other limit? Suppose you have chosen some number of sides, and their lengths can vary. If you have $3$ sides and they have lengths of $1$, you have an equilateral triangle with side lengths of $1$. For any finite $l$, you will have equilateral triangle with side length $l$. Now if you suppose you have a tiny tiny polygon and it's still a triangle, the side length is clearly some finite $l > 0$. Hence $\lim_{l\rightarrow0}P(n, l)$ must be a single point. No matter how many "sides" you give to a single point it will always just be a single point. Therefore,
\begin{align*}
\lim_{n\rightarrow\infty}\lim_{l\rightarrow0}P(n, l) = \lim_{n\rightarrow\infty}(\lim_{l\rightarrow0}P(n, l)) = \lim_{n\rightarrow\infty}(\mathrm{Single\ point}) = \mathrm{Single\ point}
\end{align*}
Clearly a single point is not an infinite line, and they're both certainly not circles. The flaw in your argument is that "the polygon formed when side lengths go to zero and the number of lines goes to infinity" cannot be uniquely described as a circle.
Also I apologise to everybody for my loose use of the word "polygon"
For an n-gon,
considering a line
rotating around it
rotates 360.
This implies the standard result
that
the sum of the exterior angles
in the direction of the rotation
is 360
so the sum of the interior angles
is n*180-360 = 180(n-2).
Therefore
the sum of the exterior angles
on both sides of each vertex
is 720.
From your first diagram,
each angle of
the extended polygon
is 180 minus the
sum of the adjacent exterior angles,
so their sum is
180*n-720
=180(n-4).
Best Answer
Let's have a polygon. Let's draw a line at some angle $\theta_1$ to some fixed direction. By moving the line parallel to itself, we can make the whole polygon to lie either on one side of the line or the other. Thus, since the function “area on one side minus area on the another” is continuous, it should go through zero. Thus, for every $\theta_1$ there is a line $AD = L(\theta_1)$ that splits the polygon in halves.
Using the same mean value theorem we can show, that for every $\theta_1$, there is $\theta_2$ such that lines $L_1=AD=L(\theta_1)$ and $L_2=BE=L(\theta_2)$ split the area in a proportion $k=[APB]/[BPD]=1/2$. Indeed, if $\theta_2=\theta_1$, then $L_1=L_2$ and $k=0$. However, if $\theta_2=\theta_1+\pi$ (we rotated the line 2 by $\pi$ and $L_1=L_2$ again), then $k=\infty$. Given $k(\theta_2)$ is continuous, there should be some $\theta_2$, so $k=1/2$
By the same argument, for every $\theta_1$ there is always a $\theta_3$ and line $L_3=CF=L(\theta_3)$, so $k=[AQC]/[CQD]=2$.
Now consider point $R$, the intersection of $L_2$ and $L_3$. As we traverse from $\theta_1\to\theta_1+\pi$, lines $L2\leftrightarrow L3$ and $P\leftrightarrow Q$, but $R\to R$. However, now $R$ lies on the other side of line $L_1$. That means that during its journey, it crossed the line $L_1$. At this moment all three lines pass through one point.
By the way, we didn't use neither property of convexity, nor that it is a polygon. What mattered is that the area of the intersection of this shape with half-plane is continuous in respect to movements of the half-plane.