I want to check if my proof is correct.
Consider the subgroup $H$ of $G$ which generated by $U$. Then it's enough to show that $H=G$. But, if $H\neq G$, then $G$ can be written as a disjoint union of cosets $gH$ i.e. $G=\cup_{g\in G} gH$. So, it's enough to show that $gH$ is open since in that case we will write $G$ as a disjoint union of two non-empty open sets $H$ and $\cup_{g\neq e\in G}gH$ i.e. $$G=H\cup(\cup_{g\neq e\in G}gH)$$ Which contradicts the fact that $G$ is connected.
To show that $gH$ is open it's enough to show that $H$ is open since $gH$ is just the image of $L_g(H)$ where $L_g:G\to G$ is an automorphism of $G$ given by $L_g(a)=ga$. But, $H$ is open since $H=\cup_{n\in\mathbb{N}} U^n$.
Does it work?
What you have proved is that connected Lie groups have no proper, non-trivial, open subgroups. Which is true. And then applied it to the special case $H=\langle U\rangle$. I see no obvious problem with it.