Irska

Write $S=\{x_1,\dots ,x_n\}$ where $x_1<\cdots <x_n$. The $2n-3$ elements $$x_1+x_2<\cdots <x_1+x_n<x_2+x_n<x_3+x_n<\cdots <x_{n-1}+x_n$$ account for all the elements of $T$. Now $x_2+x_{n-1}$ lies between $x_1+x_{n-1}$ and $x_2+x_n$, so it must be $x_1+x_n$. So $x_2-x_1=x_n-x_{n-1}=d$, say. Proceeding by induction, having shown that $x_i-x_{i-1}=d$, for some $2\le i<n-1$, observe that $x_{i+1}+x_{n-1}$ lies between $x_i+x_{n-1}=x_{i-1}+x_n$ and $x_{i+1}+x_n$, so $x_{i+1}+x_{n-1}=x_i+x_n$, giving $x_{i+1}-x_i=d$.