Does there exist a topological group where every point has a (nontrivial) neighborhood which is a subgroup? Does there exist a topological group which can be covered by (nontrivial and proper) open subgroups of itself? If so, what are groups of these types called and is this a nice property for a topological group to have? Or is this just impossible?
 A: It's certainly possible, consider $G = \mathbb{Z}^2$ with the discrete topology. If $x \in G$ is any element, consider $H$ to be the subgroup generated by $x$ (unless $x = 0$ in which case consider $H = \mathbb{Z} \times \{0\}$ for example). It's proper because $G$ has rank $2 > 1$, nontrivial, and open because $G$ is discrete.
If you want an example that's not discrete, consider $G = \mathbb{R} \times \mathbb{Z}^2$ and apply a similar reasoning. I don't know if there's a connected example and I don't know if this property has a name.
A: There's $\Bbb Q_p$, the $p$-adic numbers under addition. This is an infinite, locally compact and totally disconnected group. The subgroups
$p^n \Bbb Z_p$ for $n\in\Bbb Z$ cover $\Bbb Q_p$ and are open therein
(and are compact to book).
A: (1). An ordered group is a group $G$ with a linear order $<$ such  that $a<b\implies ((ac<bc)\land (ca<cb))$ for all $a,b,c \in G$. 
(2). Consider the free Abelian group $G$ on $\{a_n:n\in \mathbb N\}$  where $m\ne n\implies a_m\ne a_n,$ with identity element $1.$ For $1\ne x\in G$ there is a unique finite non-empty $S\subset \mathbb N$ and unique $\{e_s:s\in S\}\subset \mathbb Z$ \ $\{0\}$ such that $x=\prod_{s\in S}(x_s)^{e_s}.$ 
We can linearly order $G$ by declaring that $x_1>1$ and that $x_{n+1}>x_n^j$ for all $n, j \in \mathbb N ,$ and applying the rules of (1), whereupon $G$ is an ordered group.
Now let $G$ have the $<$-order topology. Then $G$ is a topological group.  Each $G_n=\{y\in G: x_{n+1}^{-1}<y<x_{n+1}\}$ is an open-and-closed proper subgroup of $G.$  Every $y\in G$ belongs to some $G_n.$
Remark. It is not necessary to require, in this example, that $G$ be Abelian. It just enables us to  skip some intermediate results that would be needed to justify  the claims that $G$ is an ordered group and that $G_n$ is an open  proper subgroup.
