Alternative approach: Inclusion Exclusion
Much less elegant, but still viable.
See this article for an
introduction to Inclusion-Exclusion.
Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.
Let $~S~$ denote the set of all non-empty subsets of $~\{1,2,\cdots,8\} \implies |S| = \displaystyle \left(2^8\right) - 1.$
For $~k \in \{1,2,\cdots,7\},~$ let $~S_k~$ denote the subset of $~S~$ that contains both of the elements $~\{k,k+1\}.$
Then, the desired enumeration is
$$|S| - |S_1 \cup S_2 \cup \cdots \cup S_7|.$$
Let $~T_0~$ denote $~|S| \implies T_0 = \displaystyle \left(2^8\right) - 1 = 255. \tag1 $
Let $~T_1~$ denote $~\displaystyle \sum_{k=1}^7 |S_k|.~$
That is, $~T_1~$ represents the sum of $~\displaystyle \binom{7}{1}~$ numbers.
For $~r \in \{2,3,4,5,6,7\},~$ let $~T_r~$ denote
$\displaystyle \sum_{1 \leq i_1 < i_2 < \cdots < i_r \leq 7} |S_{i_1} \cap S_{i_2} \cap \cdots \cap S_{i_r}|.$
That is, $~T_r~$ represents the sum of $~\displaystyle \binom{7}{r}~$ numbers.
Then, in accordance with Inclusion Exclusion Theory,
$$|S| - |S_1 \cup S_2 \cup \cdots \cup S_7| =
\sum_{r = 0}^7 (-1)^r T_r. \tag2 $$
Therefore, the entire problem reduces to computing $~T_r ~: r \in \{1,2,\cdots,7\},~$ and then performing the computation on the RHS of (2) above.
$\underline{\text{Computation of} ~T_1}$
For a set that is part of the collection represented by $~S_1~$, the set must contain the specific elements $~1~$ and $~2.~$ Therefore, $~|S_1| = 2^6 = 64.~$
By symmetry, for each $~k \in \{2,3,4,5,6,7\},~$ you have that $~|S_k| = |S_1|.~$
Therefore,
$$~T_1 = 7 \times 64 = 448. \tag3 $$
$\underline{\text{Computation of} ~T_2}$
For a set that is part of the collection represented by $~S_1\cap S_2~$, the set must contain the specific elements $~1, ~2,~$ and $~3.$ Therefore, $~|S_1 \cap S_2| = 2^5 = 32.~$
For a set that is part of the collection represented by $~S_1\cap S_3~$, the set must contain the specific elements $~1, ~2, ~3,~$ and $~4.$ Therefore, $~|S_1 \cap S_3| = 2^4 = 16.~$
By symmetry, when examining the set intersections that will be used to compute the $~\displaystyle \binom{7}{2} = 21~$ terms in $~T_2,~$ exactly $~6~$ of them will have an enumeration of $~32~$ and the remaining $~(21 - 6) = 15~$ terms will have an enumeration of $~16.~$
Therefore,
$$~T_2 = (6 \times 32) + (15 \times 16) = 432. \tag4 $$
$\underline{\text{Computation of} ~T_3}$
For a set that is part of the collection represented by $~S_1\cap S_2 \cap S_3~$, the set must contain the specific elements $~1, ~2, ~3,~$ and $~4.$ Therefore,
$~|S_1 \cap S_2 \cap S_3| = 2^4 = 16.~$
For a set that is part of the collection represented by $~S_1\cap S_2 \cap S_4~$, the set must contain the specific elements $~1, ~2, ~3, ~4,~$ and $~5.$ Therefore,
$~|S_1 \cap S_2 \cap S_4| = 2^3 = 8.~$
For a set that is part of the collection represented by $~S_1 \cap S_3 \cap S_5~$, the set must contain the specific elements $~1, ~2, ~3, ~4, ~5~$ and $~6.$ Therefore, $~|S_1 \cap S_3 \cap S_5| = 2^2 = 4.~$
The computation of $~T_3~$ is complicated enough that I choose to complete this section by presenting the corresponding data lexicographically, in a table.
\begin{array}{| r | r | r | r |}
\hline
\text{Initial Sets} & \text{Added Set} & \text{Terms} & \text{Total} \\
\hline
1,2 & [3 : 4 : 5 : 6 : 7] & 16 + 8 + 8 + 8 + 8 & 48 \\
1,3 & [4 : 5 : 6 : 7] & 8 + 4 + 4 + 4 & 20 \\
1,4 & [5 : 6 : 7] & 8 + 4 + 4 & 16 \\
1,5 & [6 : 7] & 8 + 4 & 12 \\
1,6 & [7] & 8 & 8 \\
\hline
2,3 & [4 : 5 : 6 : 7] & 16 + 8 + 8 + 8 & 40 \\
2,4 & [5 : 6 : 7] & 8 + 4 + 4 & 16 \\
2,5 & [6 : 7] & 8 + 4 & 12 \\
2,6 & [7] & 8 & 8 \\
\hline
3,4 & [5 : 6 : 7] & 16 + 8 + 8 & 32 \\
3,5 & [6 : 7] & 8 + 4 & 12 \\
3,6 & [7] & 8 & 8 \\
\hline
4,5 & [6 : 7] & 16 + 8 & 24 \\
4,6 & [7] & 8 & 8 \\
\hline
5,6 & [7] & 16 & 16 \\
\hline
& & & \color{red}{280} \\
\hline
\end{array}
Therefore,
$$~T_3 = 280. \tag5 $$
$\underline{\text{Subsequent} ~T_r~ \text{Computations}}$
Initially, I considered that (for example) there is a 1-1 correspondence between the $~\displaystyle \binom{7}{3} = 35~$ set intersections represented in the enumeration of $~T_3~$ and the $~\displaystyle \binom{7}{4} = 35~$ set intersections represented in the enumeration of $~T_4.$
However, consider the specific $~T_3~$ terms represented by
$~|S_1 \cap S_2 \cap S_3|,~$ and $~|S_2 \cap S_3 \cap S_4|.~$ These two terms have the same enumeration.
Now consider the corresponding $~T_4~$ terms represented by
$~|S_4 \cap S_5 \cap S_6 \cap S_7|~$ and $~|S_1 \cap S_5 \cap S_6 \cap S_7|.~$ These two terms do not have the same enumeration. Therefore, I abandoned this approach.
Instead, for the computations of $~T_4, T_5,~$ and $~T_6,~$ I also choose to present the corresponding data lexicographically, in a table.
$\underline{\text{Computation of} ~T_4}$
\begin{array}{| r | r | r | r |}
\hline
\text{Initial Sets} & \text{Added Set} & \text{Terms} & \text{Total} \\
\hline
1,2,3 & [4 : 5 : 6 : 7] & 8 + 4 + 4 + 4 & 20 \\
1,2,4 & [5 : 6 : 7] & 4 + 2 + 2 & 8 \\
1,2,5 & [6 : 7] & 4 + 2 & 6 \\
1,2,6 & [7] & 4 & 4 \\
\hline
1,3,4 & [5 : 6 : 7] & 4 + 2 + 2 & 8 \\
1,3,5 & [6 : 7] & 2 + 1 & 3 \\
1,3,6 & [7] & 2 & 2 \\
\hline
1,4,5 & [6 : 7] & 4 + 2 & 6 \\
1,4,6 & [7] & 2 & 2 \\
\hline
1,5,6 & [7] & 4 & 4 \\
\hline
2,3,4 & [5 : 6 : 7] & 8 + 4 + 4 & 16 \\
2,3,5 & [6 : 7] & 4 + 2 & 6 \\
2,3,6 & [7] & 4 & 4 \\
\hline
2,4,5 & [6 : 7] & 4 + 2 & 6 \\
2,4,6 & [7] & 2 & 2 \\
\hline
2,5,6 & [7] & 4 & 4 \\
\hline
3,4,5 & [6 : 7] & 8 + 4 & 12 \\
3,4,6 & [7] & 4 & 4 \\
\hline
3,5,6 & [7] & 4 & 4 \\
\hline
4,5,6 & [7] & 8 & 8 \\
\hline
& & & \color{red}{129} \\
\hline
\end{array}
Therefore,
$$~T_4 = 129. \tag6 $$
$\underline{\text{Computation of} ~T_5}$
\begin{array}{| r | r | r | r |}
\hline
\text{Initial Sets} & \text{Added Set} & \text{Terms} & \text{Total} \\
\hline
1,2,3,4 & [5 : 6 : 7] & 4 + 2 + 2 & 8 \\
1,2,3,5 & [6 : 7] & 2 + 1 & 3 \\
1,2,3,6 & [7] & 2 & 2 \\
\hline
1,2,4,5 & [6 : 7] & 2 + 1 & 3 \\
1,2,4,6 & [7] & 1 & 1 \\
\hline
1,2,5,6 & [7] & 2 & 2 \\
\hline
1,3,4,5 & [6 : 7] & 2 + 1 & 3 \\
1,3,4,6 & [7] & 1 & 1 \\
\hline
1,3,5,6 & [7] & 1 & 1 \\
\hline
1,4,5,6 & [7] & 2 & 2 \\
\hline
2,3,4,5 & [6 : 7] & 4 + 2 & 6 \\
2,3,4,6 & [7] & 2 & 2 \\
\hline
2,3,5,6 & [7] & 2 & 2 \\
\hline
2,4,5,6 & [7] & 2 & 2 \\
\hline
3,4,5,6 & [7] & 4 & 4 \\
\hline
& & & \color{red}{42} \\
\hline
\end{array}
Therefore,
$$~T_5 = 42. \tag7 $$
$\underline{\text{Computation of} ~T_6}$
\begin{array}{| r | r | r | r |}
\hline
\text{Initial Sets} & \text{Added Set} & \text{Terms} & \text{Total} \\
\hline
1,2,3,4,5 & [6 : 7] & 2 + 1 & 3 \\
1,2,3,4,6 & [7] & 1 & 1 \\
\hline
1,2,3,5,6 & [7] & 1 & 1 \\
\hline
1,2,4,5,6 & [7] & 1 & 1 \\
\hline
1,3,4,5,6 & [7] & 1 & 1 \\
\hline
2,3,4,5,6 & [7] & 2 & 2 \\
\hline
& & & \color{red}{9} \\
\hline
\end{array}
Therefore,
$$~T_6 = 9. \tag8 $$
$\underline{\text{Computation of}}$
$T_7~$ represents $|S_1 \cap S_2 \cap \cdots \cap S_7|$ which implies that the subset must have each of the elements in $~\{1,2,3,\cdots,8\}.$
Therefore,
$$~T_7 = 1. \tag9 $$
$\underline{\text{Final Computation}}$
$$T_0 - T_1 + T_2 - T_3 + T_4 - T_5 + T_6 - T_7 = \\
(T_0 + T_2 + T_4 + T_6) - (T_1 + T_3 + T_5 + T_7) = \\
(255 + 432 + 129 + 9) - (448 + 280 + 42 + 1) = 54.$$