= Solution
Use base-two <logarithms>, so <Von Neumann entropy> is measured in bits, and set $0\log0=0$. Write $D(\rho\Vert\sigma)=\operatorname{Tr}\rho(\log_2\rho-\log_2\sigma)$ for the <quantum relative entropy>, the quantity denoted $S(\rho\Vert\sigma)$ in the paper. It is $+\infty$ unless the <support of a positive operator> $\rho$ lies within that of $\sigma$.
We first establish <nonnegativity of quantum relative entropy>, including its equality condition. Let $\rho=\sum_{i:r_i>0}r_i|u_i\rangle\langle u_i|$ and $\sigma=\sum_js_j|v_j\rangle\langle v_j|$, and suppose the support condition holds. Put $q_i=\langle u_i|\sigma|u_i\rangle>0$. Concavity of the scalar logarithm gives
$$
\langle u_i|\log\sigma|u_i\rangle=\sum_j|\langle u_i|v_j\rangle|^2\log s_j\leq\log q_i.
$$
Terms with $s_j=0$ have zero weight here. Working first with natural logarithms, the inequality $-\ln x\geq1-x$ yields
$$
(\ln2)D(\rho\Vert\sigma)\geq\sum_{i:r_i>0}r_i\ln\frac{r_i}{q_i}\geq\sum_{i:r_i>0}(r_i-q_i)\geq0,
$$
since the $q_i$ sum to at most $\operatorname{Tr}\sigma=1$. Equality forces $q_i=r_i$ for every positive-weight eigenvector, no remaining weight of $\sigma$ outside their span, and equality in the logarithm's strict concavity. The latter makes each $u_i$ an eigenvector of $\sigma$ with eigenvalue $q_i$. Thus \b[$D(\rho\Vert\sigma)=0$ exactly when $\rho=\sigma$].
For the bipartite <density operator>,
$$
\operatorname{supp}\rho_{AB}\subseteq\operatorname{supp}\rho_A\otimes\operatorname{supp}\rho_B.
$$
Indeed, a projector onto the kernel of $\rho_A$, tensored with $I_B$, has zero expectation in $\rho_{AB}$. Positivity implies that $\rho_{AB}$ annihilates its range: $\operatorname{Tr}(\rho_{AB}P)=\|\rho_{AB}^{1/2}P\|_{\mathrm{HS}}^2=0$. The same argument applies to $B$. Hence the <quantum relative entropy> below is finite. On this support,
$$
\log(\rho_A\otimes\rho_B)=(\log\rho_A)\otimes I_B+I_A\otimes\log\rho_B.
$$
Using the <partial trace> to evaluate the two local terms gives
$$
D(\rho_{AB}\Vert\rho_A\otimes\rho_B)=S(\rho_A)+S(\rho_B)-S(\rho_{AB})\geq0.
$$
This proves <Subadditivity of Von Neumann entropy>:
$$
\boxed{S(\rho_{AB})\leq S(\rho_A)+S(\rho_B).}
$$
The proof also identifies equality exactly with the product state $\rho_{AB}=\rho_A\otimes\rho_B$.
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