Consider the 9-point affine plane $\ F_3\times F_3,\ $ over the 3-element Galois field $\ F_3.\ $ Then $\ n=9,\ $ while the respective $\ m(C)=4 < n-1$.

Here $\ C\ $ is the set of affine lines, i.e. sets described by the linear equations (homogeneous and non-homogenous, just like in an elementary school).

The explicit set of 4 straight lines which include point (0\ 0) is:

$$ \{(0\ 0)\ \ (0\ 1)\ \ (0\ 2)\}$$
$$ \{(0\ 0)\ \ (1\ 0)\ \ (2\ 0)\}$$
$$ \{(0\ 0)\ \ (1\ 1)\ \ (2\ 2)\}$$
$$ \{(0\ 0)\ \ (1\ 2)\ \ (2\ 1)\}$$

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Now let us consider a 6-point space which is a union of two parallel lines of the above plane, i.e. now $\ n=6.\ $ Then in family $\ C\ $ consists of the intersections of the straight lines with our 6-point space (all members of this new space are two 3-point sets, and nine 2-point sets). Now $\ m(C) = 4 < n-1. $ This time I believe that this $\ n=6\ $ is minimal.

Thge example is simple enough to forget about geometry--let our space be $\ \{\, 1\ 2\ 3\ 4\ 5\ 6\,\}.\ $ Then let consist of two 3-point sets:

$ \{a+1\ \ a+2\ \ a+3\}\quad $ for $\ a=0\ $ and for $\ a=3$

and nine 2-point sets:

$$ \{a\ b\}\ :\ (a\ b)\in \{1\ 2\ 3\}\times\{4\ 5\ 6\}\ $$.