For $|\mathcal{L}|=1,2,3$ one can check it. Now assuming for $|\mathcal{L}| \geq 3$, take $|\mathcal{L}|+1$ lines. By induction, $|\mathcal{L}|$ of them determine $\frac{\left(|\mathcal{L}|-1\right)\left(|\mathcal{L}|-2\right)}{2}$ regions. The $|\mathcal{L}|+1$ th intersects the other $|\mathcal{L}|$ in $|\mathcal{L}|$ points, which determine $|\mathcal{L}|-1$ segments and two half-lines on the $|\mathcal{L}|+1$ th do not. Each segment creates a new finite region, while the two half-lines do not. Hence there are $\left(|\mathcal{L}|-1\right)+\frac{\left(|\mathcal{L}|-1\right)\left(|\mathcal{L}|-2\right)}{2}=\frac{\left(|\mathcal{L}|\right)\left(|\mathcal{L}|-1\right)}{2}$ finite regions determined by $|\mathcal{L}|+1$ lines.

Inside each convex polygon, there is exactly one disc that is the inscribed disc in a triangle determined by three of its sides.

We first prove that there is a disc inside the polygon that is tangent to three of its sides (though it might not be the incircle in the triangle formed by the three sides, as the triangle formed by these might be outside the polygon). Take a small circle tangent to one of the sides $e_{1}$. Homothetise it until it touches another side $e_{2}$. Take the $A$ intersection of the lines determined by $e_{1}$ and $e_{2}$. Homothetise the disc from $A$ until it touches another side $e_{3}$ Now, take $\triangle MNP$ the triangle determined by the lines of $e_{1}, e_{2}$ and $e_{3}$. If the disc is not inscribed in $\triangle MNP$, then it is the ex-inscribed, touching side say $(NP)$ but not sides $(MN)$ and $(MP)$. Homothetise the disc from $M$, increasing its radius until it touches another side of the polygon $e_{4}$. It is now again tangent to three sides of the polygon, but its radius has increased. We can continue in this way and the radius of the disc increases. Hence at some point, the process has to stop, at which moment the disc is inscribed in the triangle determined by three sides. To prove that there is only one such disc, assume that there are two, one of which has a radius at least as big as the other. But then the one with a larger radius has to be inside the triangle of the other, and hence it has to coincide with the other disc - a contradiction.