Understanding corollary 1.5.2 Diestel Graph Theory I'm trying to Prove corollary 1.5.2 in Diestel's Graph theory (see below) but I'm having trouble wrapping my mind around the idea of a unique neighbour for every vertex.
It seems to me that a tree where the root node has four children with 4 leafs each is a counter example because all four leaves only have one neighbour (the parent). 
Can you please explain why this is not the case?
Corollary 1.5.2 The vertices of a tree can always be enumerated, say as $v_1, \cdots, v_n$ so that every $v_i$ with $i \geq 2$ has a unique neighbour in $\{v_1, \cdots, v_{i-1}\}$.
 A: If I understand correctly, "has a unique neighbor in $\{v_1, \cdots, v_{i-1}\}$" means for a vertex $v_i$, it has exactly one neighbor that is in $\{v_1, \cdots, v_{i-1}\}$ and for other neighbors of $v_i$, they are in $\{v_{i+1}, \cdots, v_n\}$.
If this is the case, it is easy to label the tree. Let $T$ be a rooted tree and $L_0, \cdots, L_h$ are sets of vertices in level $0, \cdots, h$, where $h$ is the height of the tree. To be specific, $L_0$ contains only the root $r$ of $T$ and $L_{i+1}$ contains all children of vertices in $L_{i}$.
Below is a procedure to label the vertices:

$
1.\quad i \leftarrow 0; \\
2.\quad \text{for}\ j\ \text{from}\ 0\ \text{to}\ h: \\
3.\quad\quad \text{for each vertice } v\in L_j: \\
4.\quad\quad\quad\quad \text{label }v \leftarrow v_{i+1}; \\
5.\quad\quad\quad\quad i \leftarrow i + 1; \\
6.\quad\quad \text{endfor} \\
7.\quad \text{endfor}
$

The procedure guarantees that the parent of a vertex $v$ has smaller id than $v$ and all chidlren of $v$ have greater ids.
A: I think you are misunderstanding the statement.
There exists an enumeration of the vertices such that for each $v_i$, there is one (and only one) of its neighbors within the set $\{v_1,v_2,\dots,v_{i-1}\}$.  That is not to say that for each $v_i$ there is one (and only one) neighbor in the set $\{v_1,v_2,\dots,v_{i-1},~v_{i+1},\dots,v_n\}$, neither is it saying that $v_{i-1}$ must be the neighbor of $v_i$
Take your star for example: 

If we label the vertices as pictured, note what happens.


*

*$v_2$ has exactly one neighbor in the set $\{v_1\}$ (namely $v_1$)

*$v_3$ has exactly one neighbor in the set $\{v_1,v_2\}$ (namely $v_2$)

*$v_4$ has exactly one neighbor in the set $\{v_1,v_2,v_3\}$ (namely $v_2$)

*$v_5$ has exactly one neighbor in the set $\{v_1,v_2,v_3,v_4\}$ (namely $v_2$)


We did not require that the neighbor be the one listed most recently (we don't care that $v_5$ isn't adjacent to $v_4$), just that there be exactly one neighbor in the set prior.  Indeed, for a tree, this is almost a trivial statement.  You can declare one vertex the "root", call it $r=v_1$, and then define a partial order on the vertices as: $i\leq j \Leftrightarrow dist(r,v_i)\leq dist(r,v_j)$.

Sketch of a proof:  suppose you have a tree, $T$.
Pick an arbitrary vertex and label it $v_1$.
Continue inductively, for the $i^{th}$ vertex, (with $i\geq 2$), choose it to be a neighbor of any of the first $i-1$ vertices that is not already in our list.  This process will continue until all vertices are assigned to our list.
This process will not cause a vertex to have multiple neighbors prior to it in the list.  (If that were the case, then that would imply there is a cycle and that the graph is not a tree.  Suppose $v_i$ has (at least) two neighbors $v_a$ and $v_b$ with $a<b<i$.  By construction, there is then a path from $v_b$ to $v_a$)
This process will end with all vertices in the list.  (suppose we reach a point where there is at least one unused vertex and no candidate for the new $v_i$, then that implies there are at least two connected components and the graph is not a tree).
