No. The statement only asserts (with absolute certainty) that given $n$ many $S$-neighborhood $y_1, ..., y_n$ such that $y_i \neq y_j$ for $i \neq j$, there must exists a $i$ such that $C \notin \mathcal{L}(y_i)$.
It does not say anything about collections of less than $n$ or more than $n$ $S$-neighborhood. It is consistent with the statement that there are no collection of $S$-neighborhood with the above property. There could be some $k < n$ or $k > n$ for which property holds. The only thing you can gather from the statement is that the property does not hold for exactly $n$ many $S$-neighborhood.
Also I have no idea what any of the symbols actually mean. If this is a concrete question from topology or another area of mathematics, then by using the definition of $S$-neighborhood and $\mathcal{L}(y_i)$, you may be able to get more information.I just interpreted the sentence using only its logical form.