Paracompactness of topological group. If $G$ be a locally compact topological group. Show that $G$ is paracompact.
Note: If we restrict $G$ to be locally compact, connected topological group, this problem becomes easier by constructing a sequence of sets $U_{n+1}=\bar{U}_{n}.U_1$ where $U_1$ is a neighborhood of $e$ having compact closure. However, I feel that in this proof I still have not used the full strength of local compactness (only need it to know the existence of $U_1$). In Munkres' book, he stated that if we remove the condition of connectedness, the theorem still remains true. But I have not figured it out yet. Does anyone have any idea?
 A: Firstly, most of what you are asking can be found in Hewitt's and Ross' Abstract Harmonic Analysis: Volume I, Chapter 2, (8.13).
Secondly, I think @HennoBrandsma is correct, since for a topological group $T_0$ is not only equivalent to $T_2$ (being Hausdorff), but also to regularity [2, Chapter 2, 4.8] and even complete regularity [2, Chapter 2, 8.4]. Nevertheless, it is completely fine to assume in your definition that a topological group is Hausdorff, but it is precisely the regularity condition we need in the proof, since [1, p. 163, 2.3] gives us alternative characterizations of paracompactness for regular spaces (a similar statement can be found in Munkres' book, Lemma 41.3).
We may now proceed as in [2, Chapter 2, 8.13]:
Take a symmetric neighborhood $U$ of the identity element in $G$ such that $\overline{U}$ is compact (we used local compactness here). By [2, Chapter 2, 5.7]
$$L = \bigcup_{n = 1}^\infty U^n $$
is an open and closed group (notice that if $G$ is connected, $G = L$). Once we verify that $\overline{U} \subset U^2$, we get that
$$ L = \bigcup_{n = 1}^\infty \overline{U}^n $$
is a countable union of compact spaces, so $\sigma$-compact, and hence enjoys the Lindelöf property. Since left translations are homeomorphisms, every coset $xL$ is also Lindelöf.
Now take an arbitrary open cover $\mathcal{V}$ for $G$ and let $x \in G$. Obviously $\mathcal{V}$ also covers the coset $xL$, hence there exists a finite subcover $\{V_{xL}^{(n)}\}_{n = 1}^\infty$ of $\mathcal{V}$ for $xL$.
For every $n \in \mathbb{N}$ we define
$$ \mathcal{W}_n = \{ V_{xL}^{(n)} \cap xL ; \; xL \in G/L \} $$
and we claim that
$$ \mathcal{W} = \bigcup_{n = 1}^\infty \mathcal{W}_n $$
is a refinement of $\mathcal{V}$.
Indeed, clearly for every $x \in G$ and $n \in \mathbb{N}$ we have
$$ V_{xL}^{(n)} \cap xL \subset V_{xL}^{(n)} $$
Since every $x \in G$ is contained in precisely one coset $xL \subset G$, the family $\mathcal{W}_n$ is locally finite, hence $\mathcal{W}$ is a $\sigma$-locally finite refinement of $\mathcal{V}$ that itself covers $G$.
By the above alternative characterizations of paracompactness for regular spaces, $G$ is paracompact.
Note: Since every paracompact Hausdorff space is normal, this effectively shows that every locally compact (Hausdorff) topological group is normal.
Sources:
[1] J. Dugundji, Topology, 12th printing ed., Allyn Bacon, 1978.
[2] E. Hewitt and K. A. Ross, Abstract Harmonic Analysis I, Springer-Verlag, 1979.
A: You are almost there, Let $H= \cup_n \overline U_n$, then $H$ is open and paracompact since it is a countable union of compact sets. $G$ is the union of all the cosets of $H$, this implies that $G$ is paracompact.
