 Solution

These are local criteria: multiplying by a smooth bump function extends any local test field or form to a global one without changing it near a chosen point. Assume the Lie bracket of any two sections of $T^{1,0}M$ again lies in $T^{1,0}M$. Because the almost complex structure is real, conjugating this condition shows that the Lie bracket of any two sections $U,V$ of $T^{0,1}M$ lies in $T^{0,1}M$.

For a $(1,0)$-form $\alpha$, the exterior derivative formula isThe first two terms vanish because $\alpha$ annihilates $T^{0,1}M$, and the last vanishes by bracket closure. The $(0,2)$ component is exactly the restriction of $d\alpha$ to two arguments in $T^{0,1}M$. Thus it is zero. Since the only types of a two-form are $(2,0)$, $(1,1)$ and $(0,2)$, the bracket condition implies the stated differential-form condition. This argument uses neither holomorphic coordinates nor an integrability theorem as an unproved intermediate step.