Maximal planar graph A maximal planar graph $G$ with at least 3 vertices is a simple finite planar graph for which we cannot add any new edge $e$ such that $G \cup e$ is still planar. Is there an easy and rigorous way to prove that such a graph is connected (between every 2 vertices there is a path)?
I know that such a graph is a plane triangulation. But I don't wish to use the fact that this means $G$ is 3-connected since it seems overkill. I also know there's an intuitive argument where I look at the "face with the disconnected components" and "add an edge within it connecting them". Is there a way to make this rigorous (using topological facts like in Diestel) or maybe a better proof?
 A: I think it is easy to make your intuitive argument precise.
Suppose towards a contradiction that each $G$ is a disconnected but maximal simple finite planar graph with components $G_1$ and $G_2$ for simplicity.  Since $G$ is maximal, each $G_i$ must also be maximal.  
We fix a specific planar embedding of $G$.  First, pick any planar embedding of $G_1$ and call its image $G_1'$. The boundary edges of $G_1'$ form a simple closed curve since $G_1$ is maximal, so they divide the plane into two regions, one of which is unbounded. Call this region $O_1$. Now pick a planar embedding of $G_2$ into $O_1$, and let $G'_2$ denote the image.  Define $O_2$ similarly to $O_1$ and note that $G_1' \subset 0_2$ and $G_2' \subset O_2$.         
Now since $G$ is finite, $O_1 \cap O_2$ must be path connected as each $O_i$ is $\mathbb{R}^2$ with a compact set removed. The closure of $O_1 \cap O_2$ contains a vertex $v_i$ of $G_i'$ by construction, and there is a path $e$ from $v_1$ to $v_2$.  Adding this path as an edge to $G'$ yields a planar graph, contradicting the maximality of $G$.    
