Ok, my original answer/hint was wrong and misleading, sorry about that.
The only thing I could come up with is this.
Given a set $A = \{x_1,...,x_n\} \subset R^2$ or ($R^3$) of points, consider a fixed point $x_k \in A$. On one hand, consider the trajectory:
$T_{\text{closest}} = x_k \to c_1 \to c_2 \to ... \to c_{n-1}$
($a \to b $ means that we join the points $a$ and $b$)
where $c_1$ is the closest point to $x_k$; $c_2$ is the closest point to $c_1$ that is not $x_k$; $c_3$ is the closest point to $c_2$ that is not $x_k$ and $c_1$ etc.
The length of $T_{\text{closest}}$ is
$$d(x_k,c_1)+ d(c_1,c_2) + ... + d(c_{n-2},c_{n-1}).$$
On the other hand consider the trajectory
$T_{\text{furthest}} = x_k \to f_1 \to f_2 \to ... \to f_{n-1},$
where $f_1$ is the furthest point to $x_k$; $f_2$ is the furthest point to $f_1$ that is not $x_k$; $f_3$ is the furthest point to $f_2$ that is not $x_k$ and $f_1$ etc.
Then the length of $T_{\text{farthest}}$ is
$$d(x_k,f_1)+ d(f_1,f_2) + ... + d(f_{n-2},f_{n-1}).$$
What you want to prove is:
$$d(x_k,c_1)+ d(c_1,c_2) + ... + d(c_{n-2},c_{n-1}) \leq d(x_k,f_1)+ d(f_1,f_2) + ... + d(f_{n-2},f_{n-1}) \quad (*)$$
By construction we have that $d(x_k,c_1) \leq d(x_k,f_1)$, however (as you pointed out) it is not obvious or even necessarily true that
$$d(c_i,c_{i+1}) \leq d(f_i,f_{i+1}) \; \text{for all } i \in \{1,...,n-2\}$$
Another attempt
The idea is this, I claim that for all $i \in \{1,...,n-2\}$, you can find $K \in \{1,...,n-2\}$ such that
$$d(c_i,c_{i+1}) \leq d(f_K,f_{K+1})$$
To prove this, take $d(c_i,c_{i+1})$, and take $j$ such that $f_j = c_i$, then we consider the two only possible cases.
Case 1. $c_{i+1} = f_m$ for some $j \leq m \leq n-1$. In this case, immediately we can conclude that $d(c_i,c_{i+1}) \leq d(f_j,f_{j+1})$, since $c_i = f_j$ and $c_{i+1}$ is a point not taken out from the maximal trajectory.
Case 2. $c_{i+1} = f_m$ for some $1 \leq m \leq j-1$. In this case, $c_{i+1} = f_m$ has already been taken out from the maximal trajectory, so it must happen that $d(f_m,f_j) \leq d(f_{m},f_{m+1})$, but $f_m = c_{i+1}$ and $f_j = c_i$ then we have $d(c_{i},c_{i+1}) = d(c_{i+1},c_i) \leq d(f_{m},f_{m+1})$.
In both cases we found $K \in \{1,...,n-2\}$ such that
$$d(c_i,c_{i+1}) \leq d(f_K,f_{K+1}).$$
If you can prove that this $K$ is unique for every $i$, you would be done by using the property
$$a < b \text{ and } c < d \Rightarrow a+c<b+d\text{ for all } a,b,c,d \in R$$
multiple times to prove $(*)$.
