[proofplan] We prove the interval condition for $P\times Q$ directly. Fix a strict interval between two elements of the product. If only one coordinate changes, the open interval is canonically isomorphic to an open interval in the corresponding factor, so the conclusion follows from the hypothesis. If both coordinates change, the interval is the proper part of a product of two closed bounded intervals; we use the standard product theorem for finite bounded Cohen-Macaulay posets, whose proof rests on the staircase triangulation and the fact that joins of Cohen-Macaulay complexes over $k$ are Cohen-Macaulay. [/proofplan] [step:Verify that the proposed rank function makes the product graded] The product order and the proposed rank function satisfy the hypotheses of [citetheorem:8088]. Applying that theorem to the finite graded posets $(P,\le_P)$ and $(Q,\le_Q)$ gives that $P\times Q$, ordered by $\le_{P\times Q}$, is a finite graded poset with rank function $\rho_{P\times Q}:P\times Q\to\mathbb N\cup\{0\}$ given by $\rho_{P\times Q}(p,q)=\rho_P(p)+\rho_Q(q)$. [/step] [step:Reduce Cohen-Macaulayness to an arbitrary product interval] Fix elements $(p,q),(p',q')\in P\times Q$ satisfying \begin{align*} (p,q)<_{P\times Q}(p',q'). \end{align*} We must prove that the order complex \begin{align*} \Delta\bigl(((p,q),(p',q'))_{P\times Q}\bigr) \end{align*} is Cohen-Macaulay over $k$. The inequality means that $p\le_P p'$ and $q\le_Q q'$, with at least one strict inequality. We split according to whether one or both coordinates change. [/step] [step:Handle the intervals where exactly one coordinate changes] Suppose first that $p=p'$ and $q<_Q q'$. Define \begin{align*} \varphi:(q,q')_Q&\to ((p,q),(p,q'))_{P\times Q} \end{align*} by \begin{align*} \varphi(r)=(p,r). \end{align*} This is a poset isomorphism, with inverse $(p,r)\mapsto r$. Therefore the two order complexes are isomorphic: \begin{align*} \Delta\bigl(((p,q),(p,q'))_{P\times Q}\bigr)\cong \Delta((q,q')_Q). \end{align*} By the interval Cohen-Macaulayness hypothesis on $Q$, the right-hand side is Cohen-Macaulay over $k$. The case $p<_P p'$ and $q=q'$ is identical, using the poset isomorphism \begin{align*} (p,p')_P&\to ((p,q),(p',q))_{P\times Q} \end{align*} given by \begin{align*} r&\mapsto (r,q). \end{align*} Thus every product interval in which exactly one coordinate changes is Cohen-Macaulay over $k$. This also covers rank-one intervals, because their open intervals are empty and the empty complex is Cohen-Macaulay of dimension $-1$ by convention. [guided] We first isolate the simple case where the product interval really lives in one factor. Suppose $p=p'$ and $q<_Q q'$. Then an element $(u,v)\in P\times Q$ lies strictly between $(p,q)$ and $(p,q')$ exactly when \begin{align*} (p,q)<_{P\times Q}(u,v)<_{P\times Q}(p,q'). \end{align*} The product order forces $p\le_P u\le_P p$, hence $u=p$, and it also forces $q<_Q v<_Q q'$. Therefore the open interval consists precisely of the elements $(p,v)$ with $v\in(q,q')_Q$. Define \begin{align*} \varphi:(q,q')_Q&\to ((p,q),(p,q'))_{P\times Q} \end{align*} by \begin{align*} \varphi(v)=(p,v). \end{align*} This map is order-preserving because $v\le_Q w$ implies $(p,v)\le_{P\times Q}(p,w)$. It is bijective by the preceding description of the interval, and its inverse $(p,v)\mapsto v$ is also order-preserving. Hence it is a poset isomorphism, so it induces an isomorphism of order complexes: \begin{align*} \Delta\bigl(((p,q),(p,q'))_{P\times Q}\bigr)\cong \Delta((q,q')_Q). \end{align*} The hypothesis says that every open interval in $Q$ has Cohen-Macaulay order complex over $k$, so this product interval is Cohen-Macaulay. The argument above applies verbatim when $p<_P p'$ and $q=q'$, with $P$ in place of $Q$. In rank-one cases the open interval is empty, and the statement explicitly adopts the convention that the empty complex is Cohen-Macaulay of dimension $-1$. [/guided] [/step] [step:Apply the product theorem to intervals where both coordinates change] Now suppose that \begin{align*} p<_P p' \end{align*} and \begin{align*} q<_Q q'. \end{align*} Let \begin{align*} I=[p,p']_P=\{x\in P:p\le_P x\le_P p'\} \end{align*} and \begin{align*} J=[q,q']_Q=\{y\in Q:q\le_Q y\le_Q q'\}. \end{align*} These are finite bounded graded posets, with minimum and maximum elements $p,p'$ in $I$ and $q,q'$ in $J$. Every open interval in $I$ is an open interval in $P$, and every open interval in $J$ is an open interval in $Q$. Hence $I$ and $J$ are Cohen-Macaulay over $k$ in the interval sense. We use the Product Theorem for finite bounded Cohen-Macaulay posets, in the following precise form: if $A$ and $B$ are finite bounded graded posets whose closed intervals have Cohen-Macaulay open-interval order complexes over $k$, then the order complex of the proper part of $A\times B$ is Cohen-Macaulay over $k$. Here the proper part means the product poset with only its minimum and maximum elements removed. The hypotheses of this theorem hold for $I$ and $J$. They are finite and bounded by construction, they inherit graded rank functions from $P$ and $Q$, and their interval Cohen-Macaulayness was verified above. Applying the theorem to $I$ and $J$, the order complex of $(I\times J)\setminus\{(p,q),(p',q')\}$ is Cohen-Macaulay over $k$. It remains only to identify this proper part with the desired open interval. An element $(x,y)\in I\times J$ is different from both $(p,q)$ and $(p',q')$ precisely when $(p,q)<_{P\times Q}(x,y)<_{P\times Q}(p',q')$, because $I=[p,p']_P$ and $J=[q,q']_Q$ and both coordinate inequalities are strict at the endpoints under the product order. Hence $(I\times J)\setminus\{(p,q),(p',q')\}=((p,q),(p',q'))_{P\times Q}$. Therefore $\Delta\bigl(((p,q),(p',q'))_{P\times Q}\bigr)$ is Cohen-Macaulay over $k$. [guided] The two-coordinate case is the only place where a genuinely new theorem is needed. Define $I=[p,p']_P$ and $J=[q,q']_Q$. These are finite bounded graded posets: $p$ and $p'$ are the minimum and maximum of $I$, while $q$ and $q'$ are the minimum and maximum of $J$. Every open interval in $I$ is an open interval in $P$, and every open interval in $J$ is an open interval in $Q$, so the Cohen-Macaulay interval hypotheses on $P$ and $Q$ pass to $I$ and $J$. We now apply the Product Theorem for finite bounded Cohen-Macaulay posets. The theorem says that if two finite bounded graded posets have Cohen-Macaulay open-interval order complexes over $k$, then the order complex of the proper part of their product is Cohen-Macaulay over $k$. The proper part of a bounded poset means the poset obtained by removing its minimum and maximum elements. In the present situation the minimum of $I\times J$ is $(p,q)$ and the maximum is $(p',q')$, so the proper part is $(I\times J)\setminus\{(p,q),(p',q')\}$. Finally we compare this proper part with the open interval in the original product. If $(x,y)\in I\times J$, then $p\le_P x\le_P p'$ and $q\le_Q y\le_Q q'$. Since both $p<_P p'$ and $q<_Q q'$, removing only the two endpoint pairs leaves exactly the elements strictly between $(p,q)$ and $(p',q')$ in the product order. Thus $(I\times J)\setminus\{(p,q),(p',q')\}=((p,q),(p',q'))_{P\times Q}$. The product theorem therefore gives that $\Delta\bigl(((p,q),(p',q'))_{P\times Q}\bigr)$ is Cohen-Macaulay over $k$. [/guided] [/step] [step:Conclude the interval Cohen-Macaulay condition for the product] The strict comparable pair $(p,q)<_{P\times Q}(p',q')$ was arbitrary. The preceding cases prove that its open interval has Cohen-Macaulay order complex over $k$ whether exactly one coordinate changes or both coordinates change. Hence every open interval of $P\times Q$ is Cohen-Macaulay over $k$. Together with the grading verified above, this proves that $P\times Q$, ordered componentwise and ranked by \begin{align*} \rho_{P\times Q}(p,q)=\rho_P(p)+\rho_Q(q), \end{align*} is Cohen-Macaulay over $k$ in the interval sense. [/step]