I would have expected a little more originality from a China problem The sets of teams playing "home" in each week make up a "strongly separating system of sets" for the $n$ teams (see e.g. [Bela Bollobás - The Art of Mathematics]). For $w$ weeks, the maximal number of teams is $n = \binom {w} {\lfloor w/2 \rfloor} = \binom {w} {\lceil w/2 \rceil}$, so for $w=4$ we have a maximum of $n=6$ teams (a model is $abc, ade, bdf, cef$).

A solution without using advanced basis of set theory We first prove that $7$ teams cannot finish the tournament in $4$ weeks. Let $S_i(i=1,2,\cdots ,7)$ denote the set of weeks when team $i$ play games at home. If the tournament will finish in $4$ weeks, then $S_i$s are nonempty proper subsets of $\{1,2,3,4\}$. According to rule (c), each $S_i$ is not a subset of any other $S_j$, otherwise team $i$ cannot play a home game against team $j$. On the other hand, we divide all nonempty proper subsets of $\{1,2,3,4\}$ into six sets: $A=\{\{1\},\{1,2\},\{1,2,3\}\}$, $B=\{\{2\},\{2,3\},\{2,3,4\}\}$, $C=\{\{3\},\{1,3\},\{1,3,4\}\},$ $D=\{\{4\},\{1,4\},\{1,2,4\}\}$, $E=\{\{2,4\}\}$, $F=\{\{3,4\}\}$. By the pigeonhole principle, there exist $i,j (i\not= j, i,j\in \{1,2,\cdots ,7\})$ such that $S_i, S_j$ both belong to one of those six sets listed above. That is to say $S_i\subseteq S_j$ or $S_j\subseteq S_i$ occurs. Hence $n$ can be $6$ at most. Now we design a schedule for $6$ teams satisfying the requirements: $S_1=\{1,2\}, S_2=\{1,3\}, S_3=\{1,4\}, S_4=\{2,3\}, S_5=\{2,4\}, S_6=\{3,4\}$. It is easy to verify its validity.