$ A$ is a set of points on the plane, $ L$ is a line on the same plane. If $ L$ passes through one of the points in $ A$, then we call that $ L$ passes through $ A$.

(1) Prove that we can divide all the rational points into $ 100$ pairwisely non-intersecting point sets with infinity elements. If for any line on the plane, there are two rational points on it, then it passes through all the $ 100$ sets.

(2) Find the biggest integer $ r$, so that if we divide all the rational points on the plane into $ 100$ pairwisely non-intersecting point sets with infinity elements with any method, then there is at least one line that passes through $ r$ sets of the $ 100$ point sets.
### Part (1)

1. **Define Rational Points and Sets**:
   Let \( \mathbb{Q}^2 \) denote the set of all rational points on the plane, i.e., points \((x, y)\) where both \(x\) and \(y\) are rational numbers.

2. **Partitioning Rational Points**:
   We need to partition \( \mathbb{Q}^2 \) into 100 pairwise non-intersecting sets, each containing infinitely many points. Let these sets be \( A_1, A_2, \ldots, A_{100} \).

3. **Condition on Lines**:
   We need to show that if a line \( L \) passes through two rational points, then it must intersect all 100 sets. 

4. **Proof by Contradiction**:
   Assume there exists a line \( L \) that passes through two rational points but does not intersect all 100 sets. Let \( L \) pass through points \( P_1 \in A_i \) and \( P_2 \in A_j \) where \( i \neq j \). Since \( L \) is a line through rational points, it can be described by an equation with rational coefficients.

5. **Intersection with Sets**:
   Since each \( A_k \) contains infinitely many rational points, and \( L \) is a line with rational coefficients, \( L \) must intersect each \( A_k \) at least once. This contradicts our assumption that \( L \) does not intersect all 100 sets.

6. **Conclusion**:
   Therefore, any line passing through two rational points must intersect all 100 sets.

\(\blacksquare\)

### Part (2)

1. **Objective**:
   Find the largest integer \( r \) such that if we divide all rational points on the plane into 100 pairwise non-intersecting sets with infinitely many elements, there is at least one line that passes through \( r \) of these sets.

2. **Upper Bound**:
   Consider a circle divided into 99 arcs, each containing infinitely many rational points, and the remaining points in one set. This configuration suggests that \( r \leq 3 \).

3. **Lower Bound**:
   For any four points in different sets, there is at least one intersection point. Consider this intersection point; we can find a line passing through at least three points in different sets. Thus, \( 3 \leq r \).

4. **Conclusion**:
   Combining the upper and lower bounds, we conclude that \( r = 3 \).

The final answer is \( \boxed{ r = 3 } \)