A Lie-subalgebra distribution is involutive

Choose one finite basis $E_1,\dots ,E_r$ of $\mathfrak{h}$, using the empty basis if $r=0$, and let $E_i^L(g)=d(L_g{)}_e{E}_i$. By [F1], the fields $E_1^L,\dots ,E_r^L$ form a global smooth frame for ${\mathcal{D}}^{\mathfrak{h}}$.

By [F3], $[E_i^L,E_j^L]$ is left invariant. Its value at $e$ is the Lie-algebra bracket $[E_i,E_j]$, which belongs to $\mathfrak{h}$ because $\mathfrak{h}$ is a subalgebra. If $[E_i,E_j]={\sum }_k{c}_{ij}^k{E}_k$, left invariance gives $[E_i^L,E_j^L]={\sum }_k{c}_{ij}^k{E}_k^L$, a section of ${\mathcal{D}}^{\mathfrak{h}}$.

The frame criterion [F2] applied to step 2.1 proves involutivity. When $r=0$, every local section is zero and the same conclusion is vacuous; when $r=\dim G$, the distribution is $TG$. The hypothesis [A1] is used through [F1]'s smooth tangent-bundle trivialization and [F3]'s invariant-bracket result; choosing the single finite basis in step 1.1 adds no choice.