A plane graph has finitely many faces and exactly one unbounded face

The finite union of bounded line segments lies in a sufficiently large rectangle. The exterior of that rectangle is connected and disjoint from the drawing, so it lies in one face; every unbounded face must meet the exterior and hence equals that face. Thus there is exactly one unbounded face.

Add the finitely many vertices and edge arcs one at a time. At every stage the partial drawing consists of finitely many points and straight segments by [L1]. Let $L_1,\dots ,L_s$ be the distinct full lines supporting its segments, together with a vertical line through each vertex. The complement of these lines has at most $2^s$ nonempty sign cells: for each choice of sides of the $s$ lines, the cell is an intersection of open half-planes, hence convex and connected. Every face of the drawing is a nonempty open set, so it meets the complement of this finite union of lines, which has empty interior. The sign cell containing such a point is disjoint from the drawing and connected, hence lies wholly in that face. Distinct faces contain distinct sign cells. Thus every partial drawing has at most $2^s$ faces. The empty drawing has one face, and each finite addition preserves the finite-face assertion.

Starting from the empty drawing with one face, finitely many additions yield finitely many faces by step 1.2, and step 1.1 identifies exactly one as unbounded.