A direct argument is the following. Fix a finite group $F$ of cardinality $c \geq 2$.

 1. 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 $T_2$. 

 2. The next step is to show that the canonical embedding $DL(c) \hookrightarrow T_1 \times T_2$ is quasi-isometric.
 3. 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:

 1. Notice that $F\wr \mathbb{Z}$ acts naturally on $\mathscr{W}$, and that the orbit map defines a quasi-isometric embedding $F \wr \mathbb{Z} \hookrightarrow \mathscr{W}$. 
 2. 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.
 3. 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.)