Example 3 Let $S=\{1,2, \cdots, 100\}$. Find the largest integer $k$ such that the set $S$ has $k$ distinct non-empty subsets with the property: for any two different subsets among these $k$ subsets, if their intersection is non-empty, then the smallest element in their intersection is different from the largest elements in both of these subsets. ${ }^{[2]}$
First, we prove a lemma.
Lemma If the $t_{n}$ distinct subsets of the set $T_{n}=\{1,2, \cdots, n\}$ are such that the intersection of any two subsets is non-empty, then the maximum value of $t_{n}$ is $2^{n-1}$.

Proof On one hand, the $2^{n}$ subsets of the set $T_{n}$ can be paired into $2^{n-1}$ pairs, each pair being complementary.

If $t_{n} \geqslant 2^{n-1}+1$, by the pigeonhole principle, there must be two subsets in the $2^{n}$ subsets of the set $T_{n}$ that can be paired, i.e., they are complementary. In this case, the intersection of these two subsets is the empty set, which contradicts the problem's condition.
Therefore, $t_{n} \leqslant 2^{n-1}$.
On the other hand, consider all subsets of the set $T_{n}$ that contain the element 1. There are $2^{n-1}$ such subsets, and they satisfy the condition that the intersection of any two subsets is non-empty.
In summary, the maximum value of $t_{n}$ is $2^{n-1}$.
The lemma is proved.
For the set $S_{n}=\{1,2, \cdots, n\}(n \geqslant 3)$, let the maximum number of subsets that satisfy the condition be $a_{n}$.
This problem is to find $a_{100}$.
First, consider the recursive relationship between $a_{n+1}$ and $a_{n}$.
Divide the subsets of $S_{n+1}=\{1,2, \cdots, n+1\}$ that satisfy the condition into two categories:
(1) Subsets that do not contain the element $n+1$, of which there are at most $a_{n}$.
(2) Subsets that contain the element $n+1$. For these subsets, if the element $n+1$ is removed, the remaining subsets must have non-empty intersections. According to the lemma, there are at most $2^{n-1}$ such subsets.
Thus, $a_{n+1} \leqslant a_{n}+2^{n-1}$.
Next, calculate $a_{3}$.
For the set $S_{3}=\{1,2,3\}$, divide its seven non-empty subsets into three categories:
$\{3\} 、\{1,3\} 、\{2,3\},\{2\} 、\{1,2\},\{1\} 、\{1,2,3\}$.
If $a_{3} \geqslant 4$, by the pigeonhole principle, there must be two subsets in these $a_{3}$ subsets that belong to the same category, and these two subsets do not satisfy the problem's condition. Therefore, $a_{3} \leqslant 3$.

Furthermore, take three subsets of the set $S_{3}$: $\{1,2\} 、\{1,3\}$ 、 $\{1,2,3\}$, which satisfy the problem's condition.
Thus, $a_{3}=2^{3-1}-1$.
Combining (1) and (2), for $n \geqslant 3$, we have
$$
a_{n} \leqslant a_{3}+2^{3-1}+2^{4-1}+\cdots+2^{n-2}=2^{n-1}-1 \text {, }
$$

i.e., $a_{n} \leqslant 2^{n-1}-1$.
For the set $S_{n}=\{1,2, \cdots, n\}$, the subsets that contain the element 1 and have at least two elements are exactly $2^{n-1}-1$ in number, and these $2^{n-1}-1$ subsets satisfy the problem's condition.
Therefore, $a_{n}=2^{n-1}-1$.
In this problem, $a_{100}=2^{99}-1$, i.e., the set $S=\{1,2, \cdots, 100\}$ has at most $2^{99}-1$ subsets that satisfy the problem's condition.