= Solution
For disjoint nonempty index sets $A$ and $B$, partition the mean and covariance conformably. The <conditional multivariate normal distribution> is
$$
Z_A\mid Z_B=z_B
\sim N\left(
\mu_A+\Sigma_{0,A B}\Sigma_{0,B B}^{-1}(z_B-\mu_B),
\Sigma_{0,A A}-\Sigma_{0,A B}\Sigma_{0,B B}^{-1}\Sigma_{0,B A}
\right).
$$
If the sets overlap, the shared coordinates are fixed and this formula applies to $A\setminus B$.
Partition $\Omega_0$ into the $A$ and $A^c$ blocks. The block-inverse formula identifies the displayed conditional covariance with
$$
\operatorname{Var}(Z_A\mid Z_{A^c})
=(\Omega_{0,A,A})^{-1}.
$$
For $A=\{j,k\}$, inversion of the two-by-two precision block gives
$$
\operatorname{Cov}(Z_j,Z_k\mid Z_{-jk})
=-\frac{\Omega_{0,jk}}
{\Omega_{0,jj}\Omega_{0,kk}-\Omega_{0,jk}^2}.
$$
The denominator is positive, so the conditional covariance vanishes exactly when $\Omega_{0,jk}=0$. A Gaussian pair is independent exactly when it is uncorrelated, proving
$$
Z_j\mathrel{\perp\!\!\!\perp}Z_k\mid Z_{-jk}
\quad\Longleftrightarrow\quad
\Omega_{0,jk}=0.
$$
Profiling the Gaussian likelihood over $\mu$ gives $\widehat\mu=\bar X$. Apart from constants and a positive factor, the negative log-likelihood for the <precision matrix> is
$$
-\log\det\Omega+\operatorname{tr}(S\Omega).
$$
Its derivative is $-\Omega^{-1}+S$, so the unpenalized minimizer is $S^{-1}$. The <Graphical Lasso> solves
$$
\widehat\Omega_\lambda
\in\operatorname*{argmin}_{\Omega\succ0}
\left\{-\log\det\Omega+\operatorname{tr}(S\Omega)
+\lambda\sum_{j\ne k}|\Omega_{jk}|\right\},
$$
with some conventions also penalizing the diagonal.
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