Prove that every maximal planar graph of order 4 or more is 3-connected I thought I might be able to use the fact that for a maximal planar graph the minimum degree of the  graph is at least 3, but I couldn't figure anything out. Am I headed in the right direction? Where do i go from there?
If any more detail is needed, my answer is based on this.



If an edge $e$ of a maximal planar graph is in exactly two triangles then $G/e$ is also maximal planar. (Since $G/e$ has exactly three fewer edges.)

If $G$ is a maximal planar graph with $n \geq 4$ vertices then there are at least $n$ such edges. (Induction on $n$.)

Let $G$ be a maximal planar graph with at least four vertices. Assume that there are vertices $u,v$ such that $G-\{u,v\}$ is disconnected. Let $X$ be one component of $G-\{u,v\}$ and let $Y$ be another.

Contract edges of $G$ until you're left with just $u$ and $v$ along with one vertex of $X$ and one vertex of $Y$.

Since there are now only four vertices, we must be left with $K_4$. But then removing $u$ and $v$ can't disconnect the graph, so we have a contradiction.
This is not true.  Consider $G=(V, E)$, where $V=\{1,2,3,4\}$ and $E=\{(1,2), (2,3), (3,4), (1,4), (1,3)\}$.  If we remove nodes 1 and 3, the graph is not connected, thus $G$ is not 3-connected.