A direct argument is the following. Fix a finite group $F$ of cardinality $c \geq 2$.
- Notice that the Diestel-Leader graph $DL(c)$ is the Cayley graph of $F \wr \mathbb{Z}$ with respect to some finite generating set. (See Wolfgang Woess' paper Lamplighters, Diestel-Leader graphs, random walks, and harmonic functions.)
Explicitely, $DL(c)$ is the following graph. Take two $c$-regular trees $T_1$ and $T_2$, and fix two Busemann functions $\beta_1 : T_1 \to \mathbb{R}$ and $\beta_2 : T_2 \to \mathbb{R}$. Then $DL(c)$ is the graph whose vertices are the pairs of vertices $(x,y) \in T_1 \times T_2$ satisfying $\beta_1(x)+\beta_2(y)=0$ and such that two vertices $(x_1,y_1)$ and $(x_2,y_2)$ are linked by an edge if $x_1$ and $x_2$, and $y_1$ and $y_2$, are adjacent respectively in $T_1$ and $T_2$.
- The next step is to show that the canonical embedding $DL(c) \hookrightarrow T_1 \times T_2$ is quasi-isometric.
- Finally, because a tree has Hilbert space compression one, it follows that the lamplighter group $F \wr \mathbb{Z}$ has also Hilbert space compression one.
There is another approach in my PhD thesis, but which is essentially equivalent to the previous one. Define the graph of wreaths $\mathscr{W}$ as follows:
A vertex of $\mathscr{W}$ is an equivalence class $[(\varphi,a,b)]$, where $a<b$ are integers and $\varphi \in F^{(\mathbb{Z})}$, with respect to the relation: $(\varphi_1,a_1,b_1) \sim (\varphi_2,a_2,b_2)$ if $a_1=a_2$, $b_1=b_2$ and $\varphi_1=\varphi_2$ outside $[a_1,b_1]=[a_2,b_2]$. An edge of $\mathscr{W}$ links $[(\varphi,a,b)]$ to $[(\varphi,a \pm 1,b)]$ or $[(\varphi,a,b \pm 1)]$ (if these vertices are well-defined).
Now, the point is that $\mathscr{W}$ is a median graph, or if you prefer, it is naturally the one-skeleton of a CAT(0) square complex. Next, argue as follows:
- Notice that $F\wr \mathbb{Z}$ acts naturally on $\mathscr{W}$, and that the orbit map defines a quasi-isometric embedding $F \wr \mathbb{Z} \hookrightarrow \mathscr{W}$.
- Notice also that there are two (distinct) kinds of hyperplanes in $\mathscr{W}$: the left hyperplanes, dual to edges adding $\pm 1$ to the second coordinate; and the right hyperplanes, dual to edges adding $\pm 1$ to the third coordinate. Moreover, two transverse hyperplanes cannot be both left or both right. A consequence of the previous point is that, if $T_{l}$ (resp. $T_r$) denotes the CAT(0) cube complex obtained by cubulating $\mathscr{W}$ with respect to the collection of left (resp. right) hyperplanes, then it is a tree.
- There are canonical maps $\mathscr{W} \to T_l$ and $\mathscr{W} \to T_r$: just send a vertex of $\mathscr{W}$ to the principal ultrafilter it defines. The elementary but fundamental remark is that the induced map $\mathscr{W} \to T_l \times T_r$ is an isometric embedding.
So finally the conclusion is the same: the lamplighter group $F \wr \mathbb{Z}$ quasi-isometrically embeds into a product of two trees, and therefore it must have Hilbert space compression one.
Edit: The fact that the Hilbert space compression of a tree is one follows from Guentner and Kaminker's article Exactness and uniform embeddability of discrete groups. (The argument is generalised in Campbell and Niblo's paper Hilber space compression and exactness of discrete groups to finite-dimensional CAT(0) cube complexes.)