= Solution
Use <natural units> and the <Minkowski metric> $g=\operatorname{diag}(1,-1,-1,-1)$. Multiplication of the block <gamma matrices> gives
$$
(\gamma^0)^2=I_4,\qquad
\gamma^j\gamma^k=-\begin{pmatrix}\sigma_j\sigma_k&0\\0&\sigma_j\sigma_k\end{pmatrix},\qquad
\gamma^0\gamma^j=-\gamma^j\gamma^0.
$$
The <Pauli matrices> obey $\sigma_j\sigma_k+\sigma_k\sigma_j=2\delta_{jk}I_2$, so the spatial anticommutators are $-2\delta_{jk}I_4$, the mixed ones are zero, and the temporal one is $2I_4$. Thus
$$
\boxed{\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}I_4.}
$$
Because <partial derivatives> commute, only the symmetric part of $\gamma^\mu\gamma^\nu$ survives in their contraction. Multiplying the <Dirac equation> on the left by its opposite-mass factor therefore gives the <flat-space Dirac factorization>
$$
(i\gamma^\mu\partial_\mu+m)(i\gamma^\nu\partial_\nu-m)\psi
=-(g^{\mu\nu}\partial_\mu\partial_\nu+m^2)\psi=0.
$$
Hence \b[every component satisfies $\boxed{(\Box+m^2)\psi=0}$], the <Klein-Gordon equation>. This implication does not remove the first-order <spinor> constraints imposed by the <Dirac equation>.
For <Lorentz covariance>, transform coordinates by $x'=Lx$ and the <Dirac spinor> by $\psi'(x')=S(L)\psi(x)$. The <chain rule> gives $\partial'_\mu=(L^{-1})^\nu{}_{\mu}\partial_\nu$. Consequently
$$
\begin{aligned}
S^{-1}(i\gamma^\mu\partial'_\mu-m)\psi'(x')
&=\left[iS^{-1}\gamma^\mu S(L^{-1})^\nu{}_{\mu}\partial_\nu-m\right]\psi(x)\\
&=(i\gamma^\rho\partial_\rho-m)\psi(x)=0.
\end{aligned}
$$
The cancellation uses $L^\mu{}_{\rho}(L^{-1})^\nu{}_{\mu}=\delta^\nu{}_{\rho}$. Thus the <Lorentz transformation> maps every solution to another, establishing the <Lorentz covariance of the Dirac operator>.
The infinitesimal identity follows directly from the <Clifford algebra>. Moving $\gamma^\rho$ past the two factors gives
$$
[\gamma^\mu\gamma^\nu,\gamma^\rho]
=2g^{\nu\rho}\gamma^\mu-2g^{\mu\rho}\gamma^\nu.
$$
Subtracting the same identity with $\mu,\nu$ interchanged proves the <spinor Lorentz generator commutator>
$$
[\sigma^{\mu\nu},\gamma^\rho]=2i(g^{\nu\rho}\gamma^\mu-g^{\mu\rho}\gamma^\nu).
$$
Contracting with the antisymmetric tensor $\omega_{\mu\nu}$ yields
$$
\frac{i}{4}[\sigma^{\mu\nu}\omega_{\mu\nu},\gamma^\rho]
=-\frac12\left(g^{\nu\rho}\omega_{\mu\nu}\gamma^\mu-g^{\mu\rho}\omega_{\mu\nu}\gamma^\nu\right)
=\boxed{\omega^\rho{}_{\nu}\gamma^\nu}.
$$
It also verifies $S^{-1}\gamma^\rho S=\gamma^\rho+\omega^\rho{}_{\nu}\gamma^\nu+O(\omega^2)$ for the infinitesimal <Spinor representation of the Lorentz group>.
For a positive-energy <four-momentum>, put $\psi=u(p)e^{-ip\cdot x}$, $E=\sqrt{\mathbf p^2+m^2}>0$. The <Dirac equation> becomes $(\not p-m)u=0$, where the <Feynman slash notation> uses $\not p=\gamma^\mu p_\mu=E\gamma^0-\boldsymbol\gamma\cdot\mathbf p$. Splitting $u=(\xi,\eta)^T$ gives
$$
(E-m)\xi-(\boldsymbol\sigma\cdot\mathbf p)\eta=0,\qquad
(\boldsymbol\sigma\cdot\mathbf p)\xi-(E+m)\eta=0.
$$
The second equation sets $\eta=(\boldsymbol\sigma\cdot\mathbf p)\xi/(E+m)$, and the first is then the <mass shell>, since $(\boldsymbol\sigma\cdot\mathbf p)^2=\mathbf p^2I_2$. Choosing two orthonormal two-spinors $\chi_s$ gives the <Dirac plane waves in the standard representation>:
$$
\boxed{\psi_s^{(+)}(x)=u_s(p)e^{-ip\cdot x},\qquad
u_s(p)=\sqrt{E+m}\begin{pmatrix}\chi_s\\\dfrac{\boldsymbol\sigma\cdot\mathbf p}{E+m}\chi_s\end{pmatrix},\quad s=1,2.}
$$
The negative-frequency solutions are independently
$$
\boxed{\psi_s^{(-)}(x)=v_s(p)e^{ip\cdot x},\qquad
v_s(p)=\sqrt{E+m}\begin{pmatrix}\dfrac{\boldsymbol\sigma\cdot\mathbf p}{E+m}\eta_s\\\eta_s\end{pmatrix},\qquad(\not p+m)v_s=0,}
$$
with orthonormal $\eta_s$. They are the <antiparticle> modes after field quantization. The normalization gives $u_r^\dagger u_s=v_r^\dagger v_s=2E\delta_{rs}$ and $\overline u_ru_s=2m\delta_{rs}$, $\overline v_rv_s=-2m\delta_{rs}$, using the <Dirac adjoint> $\overline u=u^\dagger\gamma^0$.
For a spatial rotation, $\sigma^{ij}=\epsilon_{ijk}\operatorname{diag}(\sigma_k,\sigma_k)$. Hence the intrinsic rotation generators are
$$
J_k=\frac12\begin{pmatrix}\sigma_k&0\\0&\sigma_k\end{pmatrix},\qquad
[J_i,J_j]=i\epsilon_{ijk}J_k,\qquad \sum_kJ_k^2=\frac34I_4.
$$
For a massive particle in its rest frame, the two independent <spinors> therefore carry the two-dimensional <spin one-half> representation, with $J_3$ <eigenvalues> $\pm1/2$ and Casimir $s(s+1)=3/4$. A $2\pi$ rotation acts as $-I$, the <spinor sign under a full spatial rotation>. Thus \b[the particle has <spin> $\boxed{s=1/2}$]; four <spinor> components encode particle and antiparticle sectors, not four <spin> states of one particle. In the <massless limit> the corresponding physical labels are <helicities> $\pm1/2$ rather than a rest-frame <spin> basis.
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