Example 4 The family of sets $\Omega$ consists of 11 five-element sets $A_{1}, A_{2}$, $\cdots, A_{11}$, where the intersection of any two sets is not empty. Let $A=\bigcup_{i=1}^{11} A_{i}=\left\{x_{1}, x_{2}, \cdots, x_{n}\right\}$, for any $x_{i} \in A$, the number of sets in $\Omega$ that contain the element $x_{i}$ is $k_{i}$, and let $m=\max \left\{k_{1}, k_{2}, \cdots, k_{n}\right\}$. Find the minimum value of $m$.
It is known that $\sum_{i=1}^{n} k_{i}=55$.
Notice that, the $k_{i}$ sets containing $x_{i}$ form
$$
\mathrm{C}_{k_{i}}^{2}=\frac{k_{i}\left(k_{i}-1\right)}{2}
$$

pairs of sets. Since the intersection of any two sets is not empty, the sum $\sum_{i=1}^{n} \mathrm{C}_{k_{i}}^{2}$ includes all pairs of sets, with some repetitions (because some pairs of sets may have more than one common element, and some elements may belong to multiple sets).

According to the condition, the 11 sets $A_{1}, A_{2}, \cdots, A_{11}$ in the family $\Omega$ have non-empty intersections, forming $\mathrm{C}_{11}^{2}=55$ unique pairs of sets.
Therefore, $\sum_{i=1}^{n} \mathrm{C}_{k_{i}}^{2} \geqslant 55$, which means
$$
\sum_{i=1}^{n} k_{i}\left(k_{i}-1\right) \geqslant 110 \text {. }
$$

Given $\max k_{i}=m$, from the above inequality we have
$$
\begin{array}{l}
110 \leqslant \sum_{i=1}^{n} k_{i}\left(k_{i}-1\right) \\
\leqslant \sum_{i=1}^{n} k_{i}(m-1)=55(m-1) .
\end{array}
$$

Thus, $m \geqslant 3$.
If $m=\max k_{i}=3$, then for any $i$, $k_{i} \leqslant 3$, which would imply that all $k_{i}=3$.

In fact, if some $k_{i} \leqslant 2$, meaning the number of sets containing element $x_{i}$ is no more than two, then at least nine sets in the family $\Omega$ do not contain $x_{i}$. Let these sets be $A_{1}, A_{2}, \cdots, A_{9}$, and let $A_{10}$ contain $x_{i}$, denoted as $A_{10}=\left\{x_{i}, a, b, c, d\right\}$. Since $A_{10} \cap A_{i} \neq \varnothing$, then
$$
A_{10} \cap A_{i} \subset\{a, b, c, d\}(i=1,2, \cdots, 9) .
$$

Therefore, among the four elements $a, b, c, d$, there must be one element (say $a$) that belongs to at least three sets in the family $\Omega$, i.e., it belongs to at least four sets among $A_{1}, A_{2}, \cdots, A_{9}, A_{10}$, which contradicts $\max k_{i}=3$.
Thus, all $k_{i}=3$.
However, this leads to $55=\sum_{i=1}^{n} k_{i}=3 n$, which is a contradiction.
Therefore, $m>3$, i.e., $m \geqslant 4$.
Furthermore, $m=4$ is achievable.
For this, a geometric construction method is used.
The "five-element set" is associated with the pentagram in Figure 3, which has 10 vertices and the midpoints of each side, totaling 15 points, labeled as 1, 2, ..., 15.
Figure 3
Construct 11 five-element sets of $A=\{1,2, \cdots, 15\}$ as follows: the 5 points on the circumference of the pentagram, the 5 points on each of the 5 sides, and the 5 points on the "cross" shapes like $(1,4,8,6,11)$, each forming a set, resulting in 11 sets.
Therefore, the minimum value of $m$ is 4.