= Solution
For a free massless <Weyl spinor>, yes: each component obeys the massless <Klein-Gordon equation>. In $c=\hbar=1$ units, take either chiral <Weyl equation>, for example
$$
(\partial_t-\boldsymbol\sigma\cdot\nabla)u=0.
$$
Apply its opposite-sign partner. The <Pauli matrix multiplication identity> makes the antisymmetric part vanish against the symmetric pair of spatial derivatives, so
$$
(\partial_t+\boldsymbol\sigma\cdot\nabla)
(\partial_t-\boldsymbol\sigma\cdot\nabla)u
=(\partial_t^2-\nabla^2)u=0.
$$
Thus $\boxed{\partial_\mu\partial^\mu u=0}$ componentwise. This does not turn the field into a scalar: its components still mix under a <Spinor representation of the Lorentz group>. The second-order equation is a necessary consequence of the first-order equation, but does not by itself impose the chiral helicity relation.
There is no unconditional massless-wave conclusion from merely being a two-component spinor, or from the null bilinear of part <4/d>. For example, $u(t,\mathbf x)=e^{-i\omega t}(1,0)^T$ has a null bilinear at every point, yet $(\partial_t^2-\nabla^2)u=-\omega^2u\ne0$ for $\omega\ne0$. Massive free two-component chiral fields coupled through a mass term instead obey the massive <Klein-Gordon equation>; interacting fields can have additional terms. The affirmative result therefore uses the free, massless dynamics.
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