Is a manifold paracompact? Should it be? We will say that a Hausdorff topological space $X$ is a smooth manifold if there is an open cover $(U_{\alpha})$ of $X$ and a corresponding collection of homeomorphisms $\varphi_{\alpha} : U_{\alpha} \to V_{\alpha} \subset \mathbb{R}^n$ such that on any overlap $U_{\alpha} \cap U_{\beta}$, the maps $$\varphi_{\beta} \circ \varphi_{\alpha}^{-1} : \varphi_{\alpha}(U_{\alpha} \cap U_{\beta}) \longrightarrow \varphi_{\beta}(U_{\alpha} \cap U_{\beta})$$ are smooth. 
Note that I have omitted the assumption of paracompactness. I recall vaguely from my undergraduate years of my lecturer telling us that there are good reasons for omitting paracompactness from the definition of a manifold unless one wants to look at metric properties of $X$ or integrate. 
His expertise was in complex geometry and homogenous spaces. I vaguely recall his justification for this being that every group can be declared a Lie group if one allows the more general definition since one is permitted to have an uncountable number of connected components. 
I am not sure if I have correctly remembered this, but I was hoping someone could maybe elaborate on this thought further? 
 A: Another good gift of paracompactness:

A Hausdorff $C^k$ manifold ($k\ge0$) is metrizable iff it is
  paracompact.

This is also true for infinite dimensional manifold modelled on a Banach space (because the underlying fact is that a topological space is metrisable iff it is Hausdorff, paracompact and locally metrisable: local metrics may be glued by means of a partition of unity).   
Talking about Banach manifolds, recall, just in case, that 

A second countable, regular, Banach manifold is paracompact. 

But here  Hausdorff  in place of regular  only gives an equivalent statement in finite dimension: it would be false for infinite dimensional Banach manifolds. (An enjoyable account of this in in R.Palais' paper Critical point theory and the minimax principle)
A: 
every group can be declared a Lie group if one allows the more general definition since one is permitted to have an uncountable number of connected components. 

A manifold is paracompact if and only if all of its connected components are second countable.
So in particular, any discrete group is a paracompact Hausdorff smooth manifold.
This manifold is second countable if and only if the group is countable,
so if we want uncountable discrete groups to be Lie groups,
we cannot require manifolds to be second countable, only paracompact.
On the other hand, removing the weaker assumption of paracompactness from the definition
of a smooth manifold will immediately eliminate all theorems
that use partitions of unity (examples: existence of Riemannian metrics, existence of connections, existence of embeddings in R^n, the Serre–Swan theorem),
which in the context of differential geometry
means the vast majority of nontrivial theorems.
Unless one intends to make all these theorems false,
it probably makes sense to make manifolds paracompact by definition.
