Every plane triangulation with at least three vertices is connected

Suppose for contradiction that $G$ is disconnected. By [L1], choose one connected component $C$ and let $H$ be the plane subgraph formed by all the other components. Both $C$ and $H$ are nonempty.

The point set of the drawing of $C$ is connected: graph paths join its vertices, their polygonal images are connected, and every point of an edge is joined along that edge to an endpoint. This drawing is disjoint from $H$, so it lies in one face $h$ of $H$.

Choose points $p_0$ in the drawing of $C$ and $x$ in the drawing of $H$. On the segment from $p_0$ to $x$, let $x_0$ be the first point in the closed drawing of $H$. The part before $x_0$ lies in the complement of $H$ and is connected to $p_0$, so it lies in $h$ and $x_0$ lies in the frontier of $h$. If $x_0$ lies in an edge interior, [L4] puts the whole edge in that frontier; hence in every case the frontier contains a vertex $y$ of $H$. The drawing of $C$ is a closed finite union of segments disjoint from $y$, so choose a disc $D$ about $y$ that avoids $C$ and meets $H$ only in $y$ and segments incident with $y$. One of the local sectors of $D\setminus H$ lies in $h$; it lies in a face $f$ of $G$, and it approaches $y$, so $y$ belongs to the boundary subgraph of $f$.

Choose a point $q$ in that sector and a point $p$ of the drawing of $C$. The face $h$ is a connected component of the open complement of $H$, so [L3] gives a polygonal path in $h$ from $q$ to $p$. Let $z$ be its first point on the closed drawing of $C$. The part before $z$ avoids every component of $G$ and starts in $f$, so it lies in $f$ and shows that $z$ is in the frontier of $f$. If $z$ is a vertex then the boundary subgraph of $f$ meets $C$ there; if $z$ is in an edge interior, [L4] puts that whole edge, and hence its endpoints in $C$, in the boundary.

The boundary subgraph of $f$ therefore contains a vertex of $C$ and a vertex of $H$. It is disconnected: any path in that boundary subgraph would also be a path in $G$ joining two distinct components, contrary to [L1].

By [F1], however, the boundary subgraph of $f$ is a triangle and is connected. This contradiction is independent of whether $h$ or $f$ is bounded, so it includes the unbounded face; it also uses only that the disconnected graph has two nonempty components, so it covers the smallest permitted order of three. Hence $G$ is connected.