What is the greatest integer $k$ which makes the statement "When we take any $6$ subsets with $5$ elements of the set $\{1,2,\dots, 9\}$,  there exist $k$ of them having at least one common element." true?

$ 
\textbf{(A)}\ 1
\qquad\textbf{(B)}\ 2
\qquad\textbf{(C)}\ 3
\qquad\textbf{(D)}\ 4
\qquad\textbf{(E)}\ 5
$
To solve this problem, we need to determine the greatest integer \( k \) such that in any selection of 6 subsets of 5 elements each from the set \(\{1, 2, \dots, 9\}\), there exist \( k \) subsets that have at least one common element.

1. **Calculate the total number of subsets:**
   The set \(\{1, 2, \dots, 9\}\) has \(\binom{9}{5}\) subsets of 5 elements. 
   \[
   \binom{9}{5} = \frac{9!}{5!(9-5)!} = \frac{9!}{5! \cdot 4!} = \frac{9 \cdot 8 \cdot 7 \cdot 6}{4 \cdot 3 \cdot 2 \cdot 1} = 126
   \]
   So, there are 126 possible subsets of 5 elements each.

2. **Apply the Pigeonhole Principle:**
   We need to select 6 subsets out of these 126. By the Pigeonhole Principle, if we distribute 30 elements (since each subset has 5 elements and there are 6 subsets, \(6 \times 5 = 30\)) among 9 possible elements, at least one element must appear in at least:
   \[
   \left\lceil \frac{30}{9} \right\rceil = 4 \text{ subsets}
   \]
   This means that there is at least one element that appears in at least 4 of the 6 subsets.

3. **Construct an example to verify:**
   We need to check if it is possible to have 6 subsets where no element appears in more than 3 subsets. If we can construct such an example, then \( k = 4 \) is the greatest integer satisfying the condition.

   Consider the following subsets:
   \[
   \{1, 2, 3, 4, 5\}, \{1, 6, 7, 8, 9\}, \{2, 3, 4, 5, 6\}, \{1, 2, 7, 8, 9\}, \{3, 4, 5, 6, 7\}, \{1, 2, 3, 8, 9\}
   \]
   - Element 1 appears in 3 subsets: \(\{1, 2, 3, 4, 5\}, \{1, 6, 7, 8, 9\}, \{1, 2, 7, 8, 9\}\)
   - Element 2 appears in 4 subsets: \(\{1, 2, 3, 4, 5\}, \{2, 3, 4, 5, 6\}, \{1, 2, 7, 8, 9\}, \{1, 2, 3, 8, 9\}\)
   - Element 3 appears in 4 subsets: \(\{1, 2, 3, 4, 5\}, \{2, 3, 4, 5, 6\}, \{3, 4, 5, 6, 7\}, \{1, 2, 3, 8, 9\}\)
   - Element 4 appears in 3 subsets: \(\{1, 2, 3, 4, 5\}, \{2, 3, 4, 5, 6\}, \{3, 4, 5, 6, 7\}\)
   - Element 5 appears in 3 subsets: \(\{1, 2, 3, 4, 5\}, \{2, 3, 4, 5, 6\}, \{3, 4, 5, 6, 7\}\)
   - Element 6 appears in 2 subsets: \(\{2, 3, 4, 5, 6\}, \{3, 4, 5, 6, 7\}\)
   - Element 7 appears in 2 subsets: \(\{1, 6, 7, 8, 9\}, \{3, 4, 5, 6, 7\}\)
   - Element 8 appears in 2 subsets: \(\{1, 6, 7, 8, 9\}, \{1, 2, 3, 8, 9\}\)
   - Element 9 appears in 2 subsets: \(\{1, 6, 7, 8, 9\}, \{1, 2, 3, 8, 9\}\)

   From this construction, we see that it is possible to have 6 subsets where no element appears in more than 4 subsets. Therefore, the greatest integer \( k \) is 4.

\(\blacksquare\)

The final answer is \( \boxed{4} \)