Example 3 Set $A=\{0,1,2, \cdots, 9\},\left\{B_{1}\right.$, $\left.B_{2}, \cdots, B_{k}\right\}$ is a family of non-empty subsets of $A$, when $i \neq j$, $B_{i} \cap B_{j}$ has at most two elements. Find the maximum value of $k$.
Solution: There are 175 subsets of one, two, and three elements, which clearly meet the requirements, so $k \geqslant 175$.

Assume $k>175$, then among $\left\{B_{1}, B_{2}, \cdots, B_{k}\right\}$, there is at least one subset with more than 3 elements. Without loss of generality, let $B_{i}$ have more than 3 elements, and $a \in B_{i}$. Denote $B_{i} \backslash\{a\}$ as the set obtained by removing $a$ from $B_{i}$, then $B_{i} \cap B_{i} \backslash\{a\}$ has at least 3 elements, so $B_{i} \backslash\{a\} \bar{E}\left\{B_{1}, B_{2}, \cdots, B_{k}\right\}$.

Now, replace $B_{i}$ with $B_{i} \backslash\{a\}$, the resulting family of subsets still meets the requirements. Continuously performing this replacement can ensure that each subset in the family has at most 3 elements, without changing the number of subsets, hence the maximum value of $k$ is 175.