Every three-connected simple graph with more than four vertices has an edge whose simple contraction remains three-connected

Suppose no edge is contractible. For every edge $xy$, the graph $G\mathrm{/}xy$ has a separator of at most two vertices. Three-connectivity of $G$ forces that separator to have the form $\{v_{xy},z\}$, where $v_{xy}$ is the contracted vertex; lifting it shows that $\{x,y,z\}$ separates $G$. Every member of this triple has a neighbour in every component of its deletion, since no proper subset can separate a three-connected graph.

Among all choices of $xy,z$ and a component $C$ of $G-\{x,y,z\}$, choose one with $\mathrm{\mid }C\mathrm{\mid }$ least, and choose a neighbour $v\in C$ of $z$. The assumed noncontractibility of $zv$ similarly gives a vertex $w$ such that $\{z,v,w\}$ separates $G$, with every member adjacent into every component of its deletion.

Because $x$ and $y$ are adjacent, some component $D$ of $G-\{z,v,w\}$ avoids both $x$ and $y$. The separator-neighbour property puts a neighbour of $v$ in $D$; since $v\in C$ and $D$ avoids $x,y,z$, that neighbour and every vertex of $D$ reached without the new separator lie in $C$. Moreover $v\mathrm{\notin }D$, so $D$ is a proper nonempty subset of $C$. The triple $\{z,v,w\}$ with component $D$ is therefore a smaller choice than $C$.

Step 3.1 contradicts the minimality in step 2.1. Hence some edge has a simple contraction that remains three-connected.