Proof. Since $G_{0}$ is a subgroup, $G=\dcups{}{i\in I}g_{i} G_{0}$ for some coset representatives $\curl{g_i}_{i}\in I$. Since left translation is a homeomorphism by Definition 3.7.3, $g_{i}G_{0}$ is homeomorphic to $G_{0}$ for each $i\in I$.

$(\implies)$ Since $G_{0}$ is a closed subspace of $G$, any open cover $\cali{U}$ of $G_{0}$ gives an open cover $\cali{V}=\cali{U}\cup \curl{G\del G_0}$ of $G$ thus if $G$ is paracompact then $G_{0}$ is paracompact.

$(\impliedby)$ Suppose $G_{0}$ is paracompact and $\cali{U}$ be an open cover of $G$. For each $i\in I$ the family

is an open cover of $g_{i}G_{0}$. Since paracompactness is a topological property, there exists a locally finite open refinement $\cali{V}_{i}$ of $\cali{U}_{i}$ for each $i\in I$. Then $\cali{V}=\cups{}{i\in I }\cali{V}_{i}$ is an open cover of $G$ that refines $\cali{U}$. Let $x\in G$ then $\exists! i\in I$ such that $x\in g_{i}G_{0}$. Since $g_{i}G_{0}$ is open, $\exists W\in\cali{N}(x)$ such that $W\suf g_{i}G_{0}$. Since $\cali{V}_{i}$ is locally finite $W$ can be taken such that $W$ intersect only finitely many sets of $\cali{V}_{i}$ and hence of $\cali{V}$.$\square$