If $G$ is a connected topological group, then every connected neighbourhood of $e$ generates $G$.

Question

Here's how I proved it:

Let $U$ be a connected neighbourhood of $e$. Then $\langle U \rangle$ is a connected subgroup of $G$ that is both open and closed. Since $G$ is connected, then $\langle U \rangle = G$.

Now, I didn't use the assumption that $U$ is connected. I feel that I must have missed something. Please enlighten me.

Answer 1

In general if $G$ is a topological group and $H\subseteq G$ is a subgroup with nonempty interior, then $H$ is both open and closed. Indeed, if $U\subseteq H$ is open and $u\in U$, then $H=\bigcup_{h\in H}hu^{-1}U$ showing it is open. And this implies that $H^c$ (the set theoretical complement of $H$) is open as well, because $H^c=\bigcup_{g\in H^c}gH$.

And so if $G$ is connected, then given any subset $A\subseteq G$ with nonempty interior we have $\langle A\rangle=G$. Therefore you are correct, connectedness of the subset is irrelevant.

Answer 2

$\langle U \rangle$ can be written as a union of sets that are continuous images of products of $U$ and that all contain $1$, so standard facts about connectedness (preserved by continuous images, products, linked unions) are needed to show it is connected. So you do use it, maybe implicitly.