Assume we have a finite set of points A in $\mathbb{R^2}$ or $\mathbb{R^3}$. Is it true that if we start from a fixed point and travel to the closest unvisited point, the sum of these distances will be no greater than the sum of lengths if we always travel to the furthest unvisited point?
It would be simple to prove it with induction, but adding or removing a point from the set $A$ can completely change the path taken, thus making it hard to use induction.
The statement sounds true, so I didn't spend much time looking for a counter-example.
Any suggestions?
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 $(*)$.