Are all minimal closed sets boxes? I'll set up a couple of (standard) definitions, then ask my question.
Definition: If $\mathbb X$ and $\mathbb Y$ are sets and $X\subseteq \mathbb X$ and $Y\subseteq \mathbb Y$, call $X{\times}Y=\{(x,y)\mid x\in X,\ y\in Y\}$ a box, and call $X$ and $Y$ the sides of the box $X{\times}Y$.
Definition: For topologies $\mathbb T_1$ and $\mathbb T_2$, the product topology $\mathbb T_1\times\mathbb T_2$ has open sets generated by boxes of open sets from $\mathbb T_1$ and $\mathbb T_2$.
Definition: Given a topology $\mathbb T$ and a point $p\in\mathbb T$, the closure of $p$, written $|p|$, is the least closed set containing $p$.
I believe it is a fact that in the product topology, the box $C_1{\times}C_2$ is closed in $\mathbb T_1\times\mathbb T_2$ if and only if its sides $C_1$ and $C_2$ are closed in $\mathbb T_1$ and $\mathbb T_2$ respectively.
My question: Given points $p_1\in\mathbb T_1$ and $p_2\in\mathbb T_2$, is the closure $|(p_1,p_2)|$ equal to the box of the closures $|p_1|{\times}|p_2|$?  It's fairly clear that $|(p_1,p_2)|\subseteq |p_1|{\times}|p_2|$, but it's less immediately evident to me whether the reverse implication must also hold, so that this would be an equality.
Proof or counterexample very welcome.  Thanks in advance.

Thanks to @Ulli for noting that this is a known result and for the precise reference.
I'll include here what seems (to me) a clean direct argument, for reference.
To prove $|(p_1,p_2)|=|p_1|{\times}|p_2|$ it would suffice to prove equality of the complements.
Say that a set $X$ avoids $x$ when $x\not\in X$, and note that $X{\times}Y$ avoids $(x,y)$ if and only if $X$ avoids $x$ and $Y$ avoids $y$.
Then the complement of $|(p_1,p_2)|$ is the union of open boxes $O_1{\times}O_2$ that avoid $(p_1,p_2)$; which is the union of open boxes $O_1{\times}O_2$ such that $O_1$ avoids $p_1$ and $O_2$ avoids $p_2$; which is the complement of $|p_1|{\times}|p_2|$.
