Clearly, any set containing only one element is a $T$-set. Also, since $\{1, 2, 3\}$ is a $T$-set, any of its subsets is certainly contained in a $T$-set.

Now let $S$ be a finite set of positive integers with at least two elements, and let $n\ (> 3)$ be the largest element in $S$. Let $\sigma(S)$ denote the sum of the elements of $S$.

Let $T_1 = \{1, 2, \dots, n\}$. Note that $\sigma(T_1) = n(n+1)/2$. Let $T_2 = T_1 \cup \{n(n+1)/2\}$. Then $\sigma(T_2) = n(n+1)$.

Finally we add another $n-2$ integers
$$(n-j)(n-j+2)(n-j+3)(n-j+4)\cdots n(n+1), \quad j=2,3,\dots,n-1$$
to obtain $T_3$. Again, it is clear that the integers are pairwise distinct and greater than $n(n+1)/2$. Now
$$\begin{aligned} \sigma(T_3) &= n(n+1) + \sum_{j=2}^{n-1} (n-j)(n-j+2)(n-j+3)(n-j+4)\cdots n(n+1) \\ &= n(n+1) + \sum_{j=2}^{n-1} [(n-j+1)-1](n-j+2)(n-j+3)(n-j+4)\cdots n(n+1) \\ &= n(n+1) + (n+1)! - n(n+1) = (n+1)! \end{aligned}$$
Hence by our construction, $T_3$ is a $T$-set, and since $S \subseteq T_3$, we are done.