Assume without loss of generality that $\ell$ is horizontal and does not pass through any vertex of the arrangement of lines in $\mathcal{L}$.

First, we shall bound the total number of edges of the upper parts of all faces intersected by $\ell$, that is, those parts that lie above $\ell$. The boundary of the upper part of such a face $K$ consists of two convex chains of edges, the left and right chain, and a portion of $\ell$. If the upper part of $K$ is bounded, then the left and right chains meet at the topmost vertex of $K$. Otherwise, the last (topmost) edges of these chains are half-lines. The edges belonging to the left (respectively, right) chain, with the exception of the topmost edge, are called the left (respectively, right) edges of $K$.
We claim that every line in $\mathcal{L}$ contains at most one left edge. Suppose, if possible, that some line in $\mathcal{L}$ has two portions $e$ and $e'$ that are left edges of $K$ and $K'$, respectively, where $e'$ is above $e$. Then the line supporting the topmost edge of the left chain of $K$ would cross $K'$, contradicting the fact that $K'$ is a face of the arrangement of lines in $\mathcal{L}$. Hence, the total number of left (respectively, right) edges of the upper parts of the faces intersected by $\ell$ is at most $|\mathcal{L}|$. Taking into account the topmost edges of the chains, the total number of edges of the upper parts of the faces intersected by $\ell$ is at most $4|\mathcal{L}|$. The same argument applies verbatim for the lower parts of these faces. Consequently, the total number of edges of the faces intersected by $\ell$ is at most $8|\mathcal{L}| - 2|\mathcal{L}| = 6|\mathcal{L}|$; the second term is due to the fact that the edges crossed by $\ell$ are counted twice: once in the upper parts and once in the lower parts.