### Part (a)

1. Initial Assumption and Setup:
Assume that no single student's list is a good list. This means that for any student $x$, there exists at least one school that is not included in $x$'s list.

2. Existence of Disjoint Lists:
Consider two students $x$ and $y$ whose lists do not share any common schools. Let $X$ and $Y$ be the sets of schools listed by $x$ and $y$, respectively. Since each list contains 12 schools and there are 2006 schools in total, it is possible to find such disjoint lists.

3. Partitioning the Lists:
Split $X$ and $Y$ into two sets of size 6 each:
$$X = A \cup C \quad \text{and} \quad Y = B \cup D$$
where $|A| = |B| = |C| = |D| = 6$.

4. Constructing Potential Good Lists:
Consider the following lists:
$$X, Y, A \cup B, A \cup D, B \cup C, C \cup D$$
We claim that at least one of these lists is a good list.

5. Contradiction Argument:
Assume for contradiction that none of these lists is a good list. This means that for each of these lists, there exists a set of 6 students such that no school in the list is included in all 6 students' lists.

6. Applying the Problem's Condition:
Take sets $S_1, S_2, S_3, S_4$ such that they are disjoint with $A \cup B, B \cup C, C \cup D, D \cup A$ respectively. According to the problem's condition, for any 6 students, there must be at least two schools that are included in the lists of all 6 students.

7. Contradiction:
Since $X$ and $Y$ are disjoint, and each $S_i$ is disjoint with one of the combined sets, it is impossible for all 6 students to avoid having at least one school from $X$ or $Y$ in their lists. This leads to a contradiction.

8. Conclusion:
Therefore, at least one of the lists $X, Y, A \cup B, A \cup D, B \cup C, C \cup D$ must be a good list.

$\blacksquare$

### Part (b)

1. Initial Setup:
Consider the total number of possible 12-element subsets of a set of 23 schools. This number is given by the binomial coefficient:
$$\binom{23}{12}$$
which is less than 2006000.

2. Constructing the Lists:
First, take $\binom{23}{12}$ students and let their lists be all possible 12-element subsets of a set of 23 schools. Then, add $2006000 - \binom{23}{12}$ other students with their lists being arbitrary subsets of the same 23 schools.

3. Ensuring No Shorter List is Good:
We need to show that no 11-element subset of schools can be a good list. Consider any 6 subsets $A, B, C, D, E, F \subset [23]$ each of size 12. Notice that any two of them must have a common element.

4. Common Element Argument:
If some four of these subsets had a common element, we would be done. Suppose $A, B, C, D$ had a common element $e$. Then, we can take the two-element subset containing $e$ and an arbitrary element in $E \cap F$.

5. Existence of Common Element:
The total number of elements in $A, B, C, D, E, F$ is $12 \times 6 = 72$. Since $72 > 3 \times 23$, by the pigeonhole principle, some element $1 \le i \le 23$ must be in at least four of the subsets $A, B, C, D, E, F$.

6. Conclusion:
This construction ensures that no 11-element subset of schools can be a good list, as required.

$\blacksquare$