Write $S = \{x_1, \dots, x_n\}$ where $x_1 < \dots < x_n$. The $2n-3$ elements
$$x_1 + x_2 < \dots < x_1 + x_n < x_2 + x_n < x_3 + x_n < \dots < x_{n-1} + x_n$$
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 $x_{i+1} - x_i = d$.