The complement of the union of all lines in $\mathcal{L}$ is the disjoint union of
$$\begin{pmatrix} |\mathcal{L}| \\ 0 \end{pmatrix} + \begin{pmatrix} |\mathcal{L}| \\ 1 \end{pmatrix} + \begin{pmatrix} |\mathcal{L}| \\ 2 \end{pmatrix}$$
open convex sets called rooms, of which exactly
$$\begin{pmatrix} |\mathcal{L}| - 1 \\ 2 \end{pmatrix}$$
are bounded. Let $\mathcal{D}_0 = \mathcal{D}_0(\mathcal{L})$ denote the set of the discs we are to count. Clearly, each disc in $\mathcal{D}_0$ is contained in some bounded room. We shall prove that each bounded contains exactly one such, whence the conclusion. To this end, we shall first prove that no bounded room contains more than one disc in $\mathcal{D}_0$, and then that each bounded room contains at least one such.

Suppose, if possible, that $R$ is a bounded room which contains two discs in $\mathcal{D}_0$; one, of radius $r$, centered at $\omega$, inscribed in the triangle $abc$ enclosed by the lines $a, b$ and $c$ in $\mathcal{L}$; and another, of radius $r'$, centered at $\omega'$, inscribed in the triangle $a'b'c'$ enclosed by the lines $a', b'$ and $c'$ in $\mathcal{L}$. Clearly, the lines $a, b, c, a', b'$ and $c'$ support edges on the boundary of $R$, the triangles $abc$ and $a'b'c'$ both contain $R$, and $\omega$ and $\omega'$ are distinct. Without loss of generality, we may assume that $r \le r'$. Since $\omega'$ lies in the interior of the triangle $abc$ and is different from $\omega$, it is within $r \le r'$ from one of the lines $a, b$ or $c$. Consequently, that line intersects the disc of radius $r'$ centered at $\omega'$ – a contradiction.

We now proceed to prove that each bounded room contains a disc in $\mathcal{D}_0$. Since $R$ is convex, and no two lines in $\mathcal{L}$ are parallel, among those lines in $\mathcal{L}$ which support edges on the boundary of $R$, at least three enclose a triangle which contains $R$ – this is a well-known fact about convex polygons different from a parallelogram. Of all such triangles, choose one with a minimal inradius. We shall prove that the open circular disc $D$ inscribed in that triangle is contained in $R$; in particular, $D$ is in $\mathcal{D}_0$. Let the triangle be enclosed by the lines $a, b$ and $c$ in $\mathcal{L}$. Since each of the lines $a, b$ and $c$ supports an edge on the boundary of $R$, it follows that $D$ and $R$ are not disjoint. Suppose, if possible, that $D$ is not contained in $R$. Then $D$ contains points on some edge on the boundary of $R$. Let $d$ be a line in $\mathcal{L}$ that supports such an edge. Clearly, $d$ is different from $a, b$ and $c$, and it is not hard to see that $d$ and two of the lines $a, b, c$ enclose a triangle which contains $R$ and has an inradius smaller than the radius of $D$ – a contradiction.