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.
Another way of saying this: for some $k$, $1 \le k \le \lfloor n/2 \rfloor$, there is only one node at distance $k$ from $s$ (otherwise, counting $s$, $t$ and at least $2$ nodes at each $k$ you'd nave at least $2 + 2 \lfloor n/2 \rfloor > n$ vertices).
Every path from $s$ to $t$ must pass through a node at distance $k$ from $s$, so if you delete the only such node you no longer have such a path.
Best Answer
The first part is correct : since the height $H$ is the distance between the root and the furthest leaf and the diameter $D$ is the largest distance between points of the graph, we have $H\leq D$.
For the second part : Let $x_1,x_2$ be any two vertices in the graph. There is a path of lenght $\leq H$ joining $x_1$ to the root and one joining $x_2$ to the root. By concatenating both, we obtained a path of length $\leq 2H$ connecting $x_1$ and $x_2$. Hence, $D\leq 2H$.