I've tried numerous ways to solve the task, but none of them worked at all. My first attempt was to try to calculate the amount of complementary events (so the sets which contain either $\{1, 3\}, \{2, 4\}, \{3, 5\}$, ... or $\{8, 10\}$). After a while of struggling, I realized I didn't really get closer to find the solution.
However, I wrote a program that counts all the possibilities, and it gave $168$ as a result - which seems about correct to me.
I’d be really grateful if any of you could help me solving this problem.
First, you can handle the odd and even numbers independently of each other, and then just multiply the number of allowed sets of odd numbers with how many allowed sets of even numbers.
Within each of these, there are only $5$ numbers to consider. In this small case it's probably easiest to just enumerate the options by hand: $1$ empty set, $5$ sets with one element, $6$ sets with two elements, $1$ set with three elements. In total $13$.
If you need to deal with a larger base set, you can also write down a recurrence. It turns out that the number of subsets of $\{1,2,3,\ldots,n\}$ that don't contain any neighboring numbers is exactly the $(n+1)$th Fibonacci number, and $F_6=13$.