= Solution
Let $\pi:T^*N\to N$ be the projection and $s_\alpha(q)=(q,\alpha_q)$ the section with image $\Gamma_\alpha$. For the <canonical one-form on a cotangent bundle>,
$$
\theta_{(q,p)}(v)=p(d\pi(v)).
$$
Since $\pi\circ s_\alpha=\mathrm{id}$, evaluation on a tangent vector $v\in T_qN$ gives
$$
(s_\alpha^*\theta)_q(v)=\alpha_q(v),\qquad s_\alpha^*\theta=\alpha.
$$
Use the same position-first <standard symplectic form> as before, $\omega=-d\theta$. Pullback commutes with the <exterior derivative>, so
$$
s_\alpha^*\omega=-d\alpha.
$$
The <graph of a differential one-form> is an embedded submanifold, because its projection is the inverse of the section on the graph. Its dimension is $\dim N$, half that of $T^*N$. Therefore the vanishing of the pulled-back symplectic form is exactly the <Lagrangian submanifold> condition, and
$$
\boxed{\Gamma_\alpha\text{ is Lagrangian}\quad\Longleftrightarrow\quad d\alpha=0}.
$$
This proves that the <graph of a closed one-form is Lagrangian> using the intrinsic one-form rather than a coordinate guess.
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