I See the following statement
let $G$ be a locally compact group with a fixed left Haar measure $\lambda$ If $G$ is not discrete , for all $x \in U$, there is an open relatively compact neighborhood $O$ of the identity element of $G$ with $O^2 \subseteq U$. We can choose a sequence $(O_n)$ of pairwise disjoint open subsets of $O$ such that $\bar O_n$ is compact and $λ(O_n) < 2^{−n}$, for all $n \geq 1$.
I'm stuck here
We can choose a sequence $(O_n)$ of pairwise disjoint open subsets of $O$ such that $\bar O_n$ is compact and $λ(O_n) < 2^{−n}$, for all $n \geq 1$
Any help will be appreciated.
Claim 1. Given an open neighborhood $O$ of the identity in a nondiscrete locally compact group, construct a sequence $O_{1},O_{2},\ldots$ of nonempty pairwise disjoint open sets contained in $O$ such that $\lambda(O_{n})<2^{-n}$.
(Note that you must want each $O_{n}$ to be nonempty or the situation is trivial. Also if you know $O$ is relatively compact then each $O_{n}$ has compact closure too.)
Proof: It suffices to fix $\epsilon>0$, and find nonempty disjoint open subsets $U$ and $V$ of $O$ such that $\lambda(U)<\epsilon$. Since if this is possible then the sets $O_{n}$ can be constructed by induction.
Nondiscreteness of $G$ means that around any point we can find an open set of arbitrarily small measure.
So start with $O$ and choose two distinct $x,y\in O$ (these exist since $O$ is not a singleton). Choose $U$ and $V$ pairwise disjoint open sets with $x\in U$ and $y\in V$. We can intersect with $O$ to assume $U$ and $V$ are subsets of $O$. By the nondiscreteness property we can further intersect $U$ with an open neighborhood of $x$ of measure less than $\epsilon$, and thus ensure $\lambda(U)<\epsilon$.