Induced subgraph is a trail of length $2r-1$ 
Let $G$ be a simple connected graph of radius $r$. Prove that we can find $2r-1$ vertices such that their induced subgraph is a trail. 

I've tried induction on $V(G)$, but haven't really made progress. The equality case occurs when $G$ is a complete graph, but I'm not sure what to make of this.
 A: FWIW, here's a solution I found to this problem.
Let $H$ be a subgraph of $G$ with the same radius but the minimal number of vertices. We show that the result holds in $H$. 
Let $v_r$ be a leaf of a spanning tree of $H$. The minimality of $H$ implies that the radius of $H-v_r$ has radius less than $r$. Thus there is a vertex $v_0$ such that $d(v_0,u)\leq r-1$ for all $u\in H-v_r$. Then $d(v_0,v_r)=r$, so let $v_0,v_1,v_2,\dots,v_r$ be the path joining $v_0,v_r$. 
Suppose that $r\geq3$ (the other case is trivial). Choose $x$ such that $d(v_2,x)=r$: then $x$ is distinct from the $v_i$'s and $d(v_0,x)\leq r-1$. 
Let $P$ be the shortest path joining $v_0$ and $x$. Let $w$ be a vertex of $P$. Suppose that $d(w,v_k)=1$ for some $k\geq2$. By the triangle inequality, $$\begin{align*}k&=d(v_0,v_k)\leq d(v_0,w)+1\\r&=d(v_2,x)\leq d(v_2,v_k)+d(v_k,w)+d(w,x)=d(w,x)+k-1.\end{align*}$$Adding these inequalities, $r\leq d(v_0,w)+d(w,x)=d(v_0,x)\leq r-1$, contradiction. Hence no $v_i$ lies on $P$, for $i\geq1$. 
Choose $w$ on $P$ adjacent to $v_1$. By the triangle inequality, $$r=d(v_2,x)\leq d(v_2,v_1)+d(v_1,w)+d(w,x)=d(w,x)+2.$$Then $x,\dots,w,v_1,v_2,\dots,v_r$ is the required path.
