= Solution
A smooth rank-$r$ <smooth distribution> $D\subset TM$ is <involutive distribution> if the <Lie bracket of vector fields> of two local sections remains a section. It is an <integrable distribution> if every point lies on an immersed $r$-dimensional <integral manifold> with tangent spaces equal to $D$. The <Frobenius theorem> says these conditions are equivalent, and gives local coordinates with $D=\operatorname{span}(\partial_1,\ldots,\partial_r)$.
For necessity, fields tangent to an <integral manifold> have brackets tangent to it: they annihilate functions vanishing on the manifold, and so does their commutator. For sufficiency, induct on $r$. The rank-zero case is immediate. Straighten a nonvanishing local section using the <flow-box theorem> to obtain $\partial_1\in D$. Choose a frame $\partial_1,Y_2,\ldots,Y_r$ with the $Y_i$ having no $\partial_1$ component. Involutivity gives $\partial_1Y=A Y$ for the column of these fields and a smooth matrix $A$. Solve the matrix ordinary differential equation $\partial_1B=-BA$, with $B=I$ on $x_1=0$. It stays invertible, and $Z=BY$ satisfies $\partial_1Z=0$. On the transverse slice, the $Z_i$ span an involutive rank-$(r-1)$ distribution. The induction hypothesis supplies adapted slice coordinates; extend them independently of $x_1$. These give the required rank-$r$ coordinate distribution and its integral manifolds.
For the <real Heisenberg group>, multiplication is
$$
(x,y,z)(a,b,c)=(x+a,y+b,z+c+xb).
$$
Differentiating left translation at the identity gives its <left-invariant frame of the real Heisenberg group>:
$$
\boxed{E_1=\partial_x,\qquad E_2=\partial_y+x\partial_z,\qquad E_3=\partial_z.}
$$
Then $[E_1,E_2]=E_3$ and the other basis brackets vanish. Since $E_3\notin\operatorname{span}(E_1,E_2)$, the <Heisenberg horizontal distribution> is not involutive, hence not integrable by the <Frobenius theorem>.
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