= Solution
Write $I=I_2$. The <Pauli matrix multiplication law> gives $\sigma^i\sigma^j+\sigma^j\sigma^i=2\delta^{ij}I$. Block multiplication of the <Weyl representation of the gamma matrices> gives
$$
(\gamma^0)^2=I_4,\qquad
\gamma^0\gamma^i=\begin{pmatrix}-\sigma^i&0\\0&\sigma^i\end{pmatrix},\qquad
\gamma^i\gamma^0=\begin{pmatrix}\sigma^i&0\\0&-\sigma^i\end{pmatrix},
$$
and
$$
\gamma^i\gamma^j=-\begin{pmatrix}\sigma^i\sigma^j&0\\0&\sigma^i\sigma^j\end{pmatrix}.
$$
Thus the mixed time-space anticommutators vanish, the time-time anticommutator is $2I_4$, and the spatial anticommutators are $-2\delta^{ij}I_4$. Together these prove the <Clifford algebra> identity
$$
\boxed{\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}I_4.}
$$
The negative spatial sign comes from the lower-left block of each spatial <gamma matrix>.
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