I encountered into the following problem from Wikipedia from "Pigeonhole principle" page.
Any subset of size six from the set $S = \{1,2,3,...,9\}$ must contain two elements whose sum is $10$.
Proof: The pigeonholes will be labelled by the two element subsets $\{1,9\}, \{2,8\}, \{3,7\}, \{4,6\}$ and the singleton $\{5\}$, five pigeonholes in all. When the six "pigeons" (elements of the size six subset) are placed into these pigeonholes, each pigeon going into the pigeonhole that has it contained in its label, at least one of the pigeonholes labelled with a two element subset will have two pigeons in it.
However I can not comprehend the last sentence of the proof. Suppose we have six element subset of set $S$, call it $\{x_1,x_2,x_3,x_4,x_5,x_6\}$. Since we have $5$ pigeonholes and $6$ pigeons (elements $x_i$) then some pigeonhole (suppose $\{1,9\}$) has at least two elements (suppose $x_1,x_2$). Why then $x_1+x_2=10$?
What if $\{x_1,x_2,x_3,x_4,x_5,x_6\}=\{1,2,4,6,8,9\}$ and for example two elements say $1,2$ are in pigeonhole $\{1,9\}$. But $1+2\neq 10$.
Can anyone explain my question in detail, please?