No small subgroups in a Lie group

Choose a smooth chart $\phi :W\to \phi (W)\subseteq {\mathbb{R}}^n$ with $\phi (e)=0$. On a smaller neighborhood of $0$, the coordinate form of the squaring map is $s(v)=\phi ({\phi }^{-1}(v{)}^2)$. The differential of multiplication at $(e,e)$ sends $(X,Y)$ to $X+Y$: its restrictions to the two coordinate axes are the identity because $ge=g$ and $eh=h$, and the differential is linear. Therefore $d{s}_0(X)=2X$.

Fix a Euclidean norm. Differentiability at $0$ gives $r>0$ such that $B_r(0)\subseteq \phi (W)$, the coordinate squaring map is defined there, and $\parallel s(v)-2v\parallel \le \frac{1}{2}\parallel v\parallel$ whenever $\parallel v\parallel <r$. Hence $\parallel s(v)\parallel \ge \frac{3}{2}\parallel v\parallel$ throughout that ball.

Put $U={\phi }^{-1}(B_r(0))$. Suppose a subgroup $K\subseteq U$ contains $h\mathrm{\ne }e$, and write $v_j=\phi (h^{2^j})$. Because every power belongs to $K\subseteq U$, all $v_j$ lie in $B_r(0)$, while $v_{j+1}=s(v_j)$. Step 2.1 gives $\parallel v_j\parallel \ge (3\mathrm{/}2{)}^j\parallel v_0\parallel \mathrm{.}$ Since $h\mathrm{\ne }e$, $\parallel v_0\parallel >0$, so the right side eventually exceeds $r$, contradicting $v_j\in B_r(0)$.

Thus every subgroup contained in $U$ is $\{e\}$. In dimension zero, the same argument reduces to the open singleton identity chart. No choice principle is used: only one chart, one norm, and one radius are fixed.