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Is there an easy proof that a non-orientable real surface $X$ does not admit a quasi-complex structure?

The proof I know is to observe that any quasi-complex structure on a real surface $X$ necessarily satisfies the integrability condition $$[T_X^{0,1}, T_X^{0,1}] \subset T_X^{0,1}$$ of the Newlander-Nirenberg theorem, because $T_X^{0,1}$ is a $1$-dimensional complex vector bundle, and the bracket $[-,-]$ is alternating, i.e. it vanishes on $T_X^{0,1}$. So by the Newlander-Nirenberg theorem, $X$ admits a complex structure, and complex manifolds have to be orientable.

However, the Newlander-Nirenberg theorem is a deep theorem and feels a bit overkill. Also I don't really see why there cannot be a quasi-complex structure. Is there a more elementary proof to convince myself?