On closed totally disconnected subgroups of connected real Lie groups  So the following statement seems to be obvious but I don't see how to prove it:
Q: How does one prove that a closed totally disconnected subgroup of a connected real Lie group is discrete?
Note that it is essential to take into account the group structure since the Cantor set is a closed totally disconnected subset of $\mathbf{R}$ which is not discrete.
 A: 0)The solution by Gleason, Montgomery, and Zippin of the fifth Hilbert's problem says that any òpcally euclidean topological group admits a compatible analytical structure.
1)By Cartan--Von Neumann theorem on closed subgroups, if $G$ is a Lie group (i.e. smooth group manifold) and $H$ is a closed subgroup of $G$, then $H$ is an embedded Lie subgroup of $G$, so its connected components w.r.t. the subspace topology and its connected components w.r.t. this Lie group structure are the same.
2)The connected components of a topological manifold are open.
Now 0),1) and 2) should be conclusive for more than your question, i.e.:

If $G$ is a locally euclidean topological group and $H$ is a totally disconnected, closed subgroup of $G$ then $H$ is discrete topological subgroup of $G$.

A: To complement what has been said earlier: a Lie group $G$ has no small subgroup, i.e. there exists a neighborhood $U$ of the identity $e\in G$ such that the only subgroup of $G$ contained in $U$, is $\{e\}$. [To see it, let $B$ be a small ball around $0$ in the Lie algebra of $G$, such that the exponential map $\exp$ induces a diffeomorphism between $B$ and its image. Set $U=\exp(B/2)$. Then for every $x\in U\backslash\{e\}$, there exists $p>1$ such that $x^p\notin U$.]
Let then $H$ be a closed subgroup of $G$. If $H$ is totally disconnected, then there is a basis of neighborhoods of $e$ in $H$, consisting of open subgroups of $H$. So $H\cap U=\{e\}$, meaning that $H$ is discrete.
A: Let's just give a quick argument:
Let $G$ be a Lie group (with Lie algebra $\mathbb g$) and $H \subset G$ be a closed subgroup. If $H$ is not discrete, then there exists a sequence $(g_n)_{n \in \mathbb N}$ of elements of $H$, which converges to $1_G$. Write $g_n = \exp( \alpha_n \xi_n)$ (for $n$ large enough), where $\xi_n \in \mathbb g$ are unit vectors (with respect to some innerproduct) and $\alpha_n \in \mathbb R_{> 0}$. Since $g_n \to 1_G$, we get $\alpha_n \to 0$. Let $\delta \in \mathbb R$ and find integers $k_n(\delta)$, such that $|\alpha_n \cdot k_n -\delta|< \alpha_n$.
Let $\xi$ be some accumulation point of the set $\lbrace \xi_n \mid n \in \mathbb N \rbrace$. Now, it is easy to see that $$g_n^{k_n(\delta)} = \exp(\alpha_n k_n(\delta) \xi_n) \to \exp(\delta \xi)$$ on a subsequence. This implies that the whole 1-parameter subgroup generated by $\xi$ lies in $H$. In particular, $H$ is not totally disconnected.
