2-Connected Graph from a Connected Graph

Question

Let $G = (V, E)$ be a connected graph with at least three vertices. Let $G'$ be the graph on V with an edge (v, w) if there is a path in G of length at most two between v and w. Show that G' is 2-connected.

My proof: For any two connected vertices v and w in G', either v or w has to be connected to another vertex, or else v and w would be an isolated component of G' (which contradicts that G' is connected since G is connected.) WLOG, let w be connected to x, then we have (v,w) and (w,x), but then we have to connect (w,x) for G'. So v, w, and x all have degree $\ge 2$. Thus G' is at least 2 connected.

Is my proof sufficient? I feel like something might be missing.

Answer 1

Degree at least $2$ does not mean $G'$ is two-connected, and the graph

v-w-x seems to give your proof problems.

Answer 2

No. Not all minimum-degree 2 graphs are 2 connected. Consider a graph consisting of 2 cycles that intersect at precisely one vertex.

HINT to show the result: Assume $G$ is not 2-connected otherwise we'd be done already. Also assume $G$ has 3 or more vertices. In fact, if $G \setminus \{v\}$ is connected so is $G' \setminus \{v\}$ So it suffices to show the following: If $v$ is such that $G \setminus \{v\}$ is disconnected then $G' \setminus \{v\}$ is connected. To this note, if $v$ has degree at least 2 in $G$, there is an edge between every pair of vertices in $N_G(v)$ [why]. [What if $v$ has degree-1 in $G$?] So this implies $G' \setminus \{v\}$ is connected [why is that].