p'_i,$ and
looking at the $\!i\!$-th coordinates of $f(p,c)$ and $f(p',c'),$
namely $(p_i,c_{m(p_i)})$ and $(p'_i,c'_{m(p_i)}),$
we see from the lexicographic ordering of $T'_i$
that the former is $>$ the latter,
giving the conclusion of~\eqref{d.notleq}.
If, on the other hand,
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.p_leq}
$p\,\leq\,p',$ \,but\, $c\,\nleq\,c',$
\end{minipage}\end{equation}
%
then in view of the second inequality above,
we may choose a $j\in n$ such that
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.c_j_notleq}
$c_j\,>\,c'_j.$
\end{minipage}\end{equation}
%
Now by the {\em first} inequality of~\eqref{d.p_leq} and
the hypothesis~\eqref{d.eq_coord}, there is some $i\in d$ such that
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.use_eq_coord}
$p_i\,=\,p'_i\in M_{m(j)}.$
\end{minipage}\end{equation}
%
Note that by~\eqref{d.f},
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.ith_coords}
$f(p,c)$ and $f(p',c')$ have $\!i\!$-th terms
$(p_i,c_j)$ and $(p'_i,c'_j)$ respectively.
\end{minipage}\end{equation}
Since the $\!T_i\!$-coordinates of the above two $\!i\!$-th terms
are the same by~\eqref{d.use_eq_coord}, the order-relation
between them is
that of the $\!C\!$-coordinates, which satisfy~\eqref{d.c_j_notleq}.
This completes the proof of~\eqref{d.notleq}, as required.
\end{proof}
Remark:
I came up with the above result after pondering~\cite{S3x2x2},
which showed by an explicit construction that
$\dim(S_3\times 2\times 2) = 3,$ i.e., is equal to $\dim(S_3).$
The next corollary includes that case.
\begin{corollary}\label{C.0&1}
Suppose $d\geq 2$ and
$(T_i)_{i\in d}$ is a family of chains, each
having a least element $0_i$ and a greatest element $1_i,$
and $P$ is a subposet of $\prod_{i\in d} T_i$ consisting of
elements each of which has at least one coordinate of
the form $0_i$ and at least one of the form $1_{i'}.$
Then for any two chains $C_0$ and $C_1,$ we have
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.dimPxC0xC1}
$\dim(P\times C_0\times C_1)\ \leq\ d.$
\end{minipage}\end{equation}
In particular, if $d\geq 3$
and $P$ is any subposet of $2^d\setminus \{0,1\},$
containing the subposet $S_d$ \textup{(}e.g., if $P=S_d),$
then for any two chains $C_0$ and $C_1,$
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.dim=d=dim}
$\dim(P\times C_0\times C_1)\ =\ d\ =\ \dim(P).$
\end{minipage}\end{equation}
%
\end{corollary}
\begin{proof}
In the context of the first assertion,
given $p\leq p'\in P,$ since $p$ has
at least one coordinate of the form $1_i,$
$p'\geq p$ must also
have $1_i$ as its $\!i\!$-th coordinate; and similarly,
since $p'$ has a coordinate $0_{i'},$ $p$ must agree with
$p'$ in that coordinate.
Applying Theorem~\ref{T.main} with $n=2,$ $M_0=\{0_i\,|\,i\in d\}$
and $M_1=\{1_i\,|\,i\in d\},$ we get~\eqref{d.dimPxC0xC1}.
The second statement follows because when each $T_i$ is
the $\!2\!$-element set $\{0_i,\,1_i\},$ the exclusion
of $0$ and $1$ from $P$ forces every element
to have at least one coordinate of
the form $1_i$ and one of the form $0_{i'};$
and the assumption that $P$ contains $S_d,$ which
we saw following~\eqref{d.S_n} has dimension $d,$ turns the
inequality~\eqref{d.dimPxC0xC1} into equality.
\end{proof}
If, instead of assuming that every element
has at least one $\!0\!$-coordinate and
at least one $\!1\!$-coordinate, we only assume one of these
conditions, it is easy to check that the analogous reasoning
gives a conclusion half as strong:
\begin{corollary}\label{C.0}
Suppose $d\geq 1,$ and $(T_i)_{i\in d}$ is a family of chains each
having a least element, $0_i,$
and $P$ is a subposet of $\prod_{i\in d} T_i$ consisting of
elements each of which has at least one coordinate of the form $0_i.$
Then for any chain $C$ we have
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.dimPxC}
$\dim(P\times C)\ \leq\ d.$
\end{minipage}\end{equation}
In particular, if $P$ is any subposet of $2^d\setminus \{1\}$
containing the subposet $S_d$ \textup{(}e.g., if $P=S_d\cup\{0\})$
then for any chain $C$ we have
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.dimPxC=}
$\dim(P\times C)\ =\ d\ =\ \dim(P).$
\end{minipage}\end{equation}
The analogous statements hold with least elements $0_i$ and $0$
everywhere replaced by greatest elements $1_i$ and $1.$\qed
\end{corollary}
To get examples of Theorem~\ref{T.main}
with $n>2,$ we shall use a different way of choosing $n$ subsets
$M_j$ of $\bigcup_{i\in d} T_i,$ based on the subscript $i$ rather
than the distinction between greatest and least elements of~$T_i.$
We will need a bit more notation.
For every positive integer $d$ and every pair of integers
$a,\,b$ with ${0\leq a< b\leq d,}$
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.Pa,b}
Let $P_d^{a,\,b}$ denote the subset of $2^d$ consisting
of those elements in which the number of coordinates
of the form $1_i$ is either $a$ or $b.$
\end{minipage}\end{equation}
%
(Thus, for $d\geq 3,$ $S_d=P_d^{1,d-1}.)$
Then we have
\begin{corollary}\label{C.Pa,a+1}
For every positive integer $d,$ every integer $a$ with
$0\leq a\leq d-1,$ and every family of chains $(C_j)_{j\in n}$ with
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.2n