Let $G$ be a connected graph with diameter $D$, and let $v$ be a vertex that is part of a diametric pair. Now define $S_i$ to be the vertices at distance $i$ from $u$. By construction, $S_i$ is not empty for $0\le i\le D$.
Now contract the $S_i$ and discard loops and multiple edges to get a graph $H$ on $D+1$ vertices $s_0, s_1,\ldots, s_D$.
Fact 1: $H$ is a path.
Proof: Build the $S_i$ via breadth-first search, which shows there is an edge between $s_i$ and $s_{i+1}$. On the other hand, shortcuts contradict the definition of $S_i$.
Now add a new edge $ij$ to $G$ to get a graph $G'$. Contract the same sets of vertices $S_i$ that were contracted to make $H$ and again discard multiple edges to get $H'$.
Fact 2: $H'$ is a path plus one edge.
Proof: The surviving edges are just the edges of $H$ plus the new edge $ij$.
Fact 3: The diameter of $H'$ is at most the diameter of $G'$.
Proof: Contracting shortens paths and discarding multiple edges leaves distances the same.
Thus if the diameter of $H'$ is at least $D/2$, then so is the diameter of $G'$, and we are done.
Edit: Pedro is right that there was a bug in my original argument for the path case. Here is a quick patch that still uses the structure of a BFS.
$H'$ has the structure of a cycle, possibly with up to two "tails" ending at $s_0$ or $s_D$; these tails meet the cycle at neighboring vertices. For zero tails, the claim is clear. For one tail, do a BFS from the end of the tail. The BFS goes one vertex per layer, splits, and then merges, so each BFS layer has at most two vertices. Again we are done.
Now look at two tails, with $a$ and $b$ non cycle vertices in each (ie the length of the tail)
and $a\le b$. Consider the BFS starting at end of the longer tail. This time, each layer has one vertex for $b$ steps, the BFS branches in a layer of two vertices, and then there are three vertices per layer for at most $a$ steps more. Since $b\ge a$, we charge the layers with three vertices to layers with one vertex, so we are done.
This is what the graph looks like in the $n-1=9$ case:
A spanning tree is a subgraph that is both (a) a tree and (b) contains every edge.
Here's an example of a spanning tree in the above graph:
There will be many others. The task set in the question is to find examples of spanning trees of diameter $d$ for $d \in \{2,3,\ldots,n-1\}$, for all $n \geq 1$.
Hint: The brown edges in the following diagram illustrate examples of spanning trees of diameter $2,\ldots,6$, respectively, in the case $n-1=6$.
I suggest you try to generalize this construction. (Note the diameter $n-2$ case is special.)
Best Answer
Graph diameter is significant in its own right, in the same way the chromatic number or maximum degree is. If you want your graph to model a network, it tells you the maximum number of 'hops' that one has to take to get from one node to any other. If your graph is embedded geometrically as, say, a graph where each edge is a straight line of length 1, then the diameter of the (abstract) graph is an upper bound on the diameter of the embedded graph, considered as a subset of $\mathbb{R}^n$.
Let $D$ be the diameter of a graph, $n$ its order, $\Delta$ its maximum degree and $\kappa$ its connectivity. Some general heuristics for how diameter behaves (these are trends, not theorems):
Finally, diameter acts as a nice complexity constraint. If you are trying to study some structure or feature in general graphs and are getting hopelessly lost, it is often useful to consider what happens in graphs of diameter 2 (especially given some other constraint or restricted graph class to go with it). Which makes it quite fortuitous that almost all graphs have diameter 2!