Not every Lie subgroup is embedded and closed

Define $i(t)=(e^{2\pi it},e^{2\pi i\alpha t})\mathrm{.}$ It is a smooth homomorphism from $(\mathbb{R},+)$ to ${\mathbb{T}}^2$. If $i(t)=(1,1)$, then $t$ and $\alpha t$ are integers, so irrationality forces $t=0$; hence $i$ is injective. Its derivative is the nonzero tangent vector $2\pi i(1,\alpha )$ at every point after translation, so it is an immersion. By [F1], its image with the transported intrinsic structure is a Lie subgroup.

This subgroup is dense by [F2], since it is the orbit through $(1,1)$ for the irrational flow. It is proper: points of the image whose first coordinate is $1$ have second coordinate in the countable set $\{e^{2\pi i\alpha n}:n\in \mathbb{Z}\}$, not all of $S^1$. Therefore the image is not closed.

It is not embedded. Fix $j\ge 1$. Irrationality makes $d_j=\min \{\parallel q\alpha \parallel :1\le q\le j\}$ positive. Choose $N$ with $1\mathrm{/}N<\min (d_j,1\mathrm{/}j)$; placing the $N+1$ fractional parts of $0,\alpha ,\dots ,N\alpha$ in $N$ equal subintervals gives, by the finite pigeonhole principle, a nonzero $q\le N$ with $\parallel q\alpha \parallel <1\mathrm{/}N$. Necessarily $q>j$. Define $q_j$ to be the least positive integer with these two properties, which makes no countable choice. Then $i(q_j)=(1,e^{2\pi i\alpha q_j})\to (1,1)$ in the ambient subspace topology, but $q_j\to +\mathrm{\infty }$ in the intrinsic copy of $\mathbb{R}$. If $i$ were an embedding, its inverse from the image to $\mathbb{R}$ would be continuous, contradicting this convergent sequence.

Thus the irrational winding is an immersed Lie subgroup that is neither closed nor embedded, refuting the claim. The intrinsic and ambient topologies, rather than the abstract subgroup law, are exactly where the failure occurs.