Group with Property (T) is Compactly Generated In this book (see page 5), it is claimed that it is straightforward to show that $G$ having property (T) implies that $G$ is compactly generated. I don't see how to prove this claim. I could use some hints. 
 A: I read that passage differently than you. What it says when I read it is that a locally compact group $G$ having property (T) is compactly generated.
Then, when I look at the table of contents on page iii, I see that section 1.3 starting on page 41 is entitled "Compact generation and other consequences", which ought to be a good place to look for a proof.
A: Here's a statement for arbitrary topological groups:
Proposition. Let $G$ be a topological group with Property T. Let $H$ be an open subgroup. Then $G$ is finitely generated over $H$ (that is, there exists a finite subset $S$ such that $G$ is generated by $H\cup S$).
Indeed, suppose the contrary. We consider the unitary representation of $G$ on $\bigoplus_L\ell^2(G/L)$, where $L$ ranges over subgroups of $G$ containing $H$ that are finitely generated over $H$. It almost has invariant vectors, in the strongest sense, that is every compact subset of $G$ fixes a unitary vector (that is, there's one vector working for all $\varepsilon$, which is usually not the case). By Property T, there is a nonzero invariant vector $f\in\bigoplus_L\ell^2(G/L)$. So $f$ has a nonzero coordinate, say $f'\in\ell^2(G/L)$ is nonzero and $G$-invariant. By the contradictory assumption, $G/L$ is infinite, so there is no nonzero invariant vector in $\ell^2(G/L)$. This is a contradiction.
Corollary: if in addition $G$ is locally compact, then it is compactly generated.
Proof: just choose $H$ compactly generated in the proposition.
(Of course this was proved in the book... you shouldn't stick to the introduction.)
