Picky note: Use curly braces for sets: $\{2,6\}$ not $(2,6)$.
You must either choose 2 from the first $\{2,6\}$ or the second $\{2,6\}$ and then are forced to choose $6$ from the opposite pair.
Next, you must choose $3$ from either $\{1,3\}$ or $\{3,4\}$.
Case 1: Choose $3$ from $\{1,3\}$ forces you to choose $1$ from $\{1,5\}$ which forces $5$ to be chosen from $\{4,5\}$ which then forces $4$ to be chosen from $\{3,4\}$
Case 2: Choose $3$ from $\{3,4\}$ forces $4$ to be chosen from $\{4,5\}$ etc.
Thus there are 2 ways to deal with 2 and 6 and 2 ways to deal with 1,3,4,5. These options are independent, so there are a total of $4=2\cdot 2$ ways to choose.