Euler's formula $\mathrm{\mid }V\mathrm{\mid }-\mathrm{\mid }E\mathrm{\mid }+\mathrm{\mid }F\mathrm{\mid }=2$ for every connected plane graph

Choose a spanning tree $T$ by [L1]. It has all $\mathrm{\mid }V\mathrm{\mid }$ vertices and $\mathrm{\mid }V\mathrm{\mid }-1$ edges, and its plane drawing has one face by the forest proposition. Hence $\mathrm{\mid }V\mathrm{\mid }-\mathrm{\mid }E(T)\mathrm{\mid }+\mathrm{\mid }F(T)\mathrm{\mid }=\mathrm{\mid }V\mathrm{\mid }-(\mathrm{\mid }V\mathrm{\mid }-1)+1=2$.

Add the edges of $G-E(T)$ in their fixed embedding. Each added edge joins vertices already connected in $T$, so it closes a cycle. By [L2] it splits one old face into two: the edge count and face count each increase by one, while the vertex count is unchanged. Thus the Euler expression remains invariant.

Equivalently, deleting a cycle edge from a connected plane graph merges its two incident faces and decreases both $\mathrm{\mid }E\mathrm{\mid }$ and $\mathrm{\mid }F\mathrm{\mid }$ by one. Deleting a bridge would disconnect the graph and is not used in this connected induction.

Starting from $T$ and adding all remaining edges yields $G$, so the invariant value from step 1.1 is $2$ for $G$.