@Edward, here is a short proof which I wrote awhile ago for my notes on group theory.
Lemma. If a 2nd countable locally compact group $G$ contains a lattice $\Gamma$ then $G$ is unimodular.
Proof. For arbitrary $g \in G$ consider the push-forward $\nu=R_g(\mu)$ of the (left) Haar measure $\mu$ on $G$; here $R_g$ is the right multiplication by $g$: $$ \nu(E)= \mu(Eg). $$ Then $\nu$ is also a left Haar measure on $G$. By the uniqueness of Haar measure, $\nu= c\mu$ for some constant $c > 0$. The lattice $\Gamma \le G$ has a fundamental domain $D\subset G$, i.e., a measurable subset of $G$ such that $$ \bigcup_{\gamma\in \Gamma} \gamma D= G, \quad \mu(\gamma D\cap D)=0, \quad \forall \gamma\in \Gamma \setminus 1. $$ In(Proof of existence of such domain uses 2nd countable assumption.) In particular, $0<\mu(D)<\infty$, since $\Gamma$ is a lattice. Then $Dg$ is again a fundamental domain for $\Gamma$ and, thus, $\mu(D)=\mu(Dg)$. Hence, $\mu(D) = \mu(Dg) = c\nu(D)$. It follows that $c=1$. Thus, $\mu$ is also a right Haar measure. qed
Another proof could be found in Chapter I of Raghunathan's book "Discrete subgroups of Lie groups".