Answer: 235

Solution:

First, the number of subsets of a block of $m$ consecutive integers containing no two consecutive elements is the Fibonacci number $F_{m+2}$ (with $F_1 = F_2 = 1$): conditioning on whether the last element is used gives the Fibonacci recursion, and the counts start $1, 2, 3, 5, \ldots$

Suppose the unique consecutive pair is $\{k, k+1\}$ for some $1 \le k \le 9.$ The remaining elements must exclude $k - 1$ and $k + 2$ (either would create a second consecutive pair) and must contain no consecutive pair within $\{1, \ldots, k-2\}$ or within $\{k+3, \ldots, 10\},$ blocks of sizes $k - 2$ and $8 - k.$ So the count for this $k$ is $F_k \cdot F_{10-k}.$

Summing over $k = 1, \ldots, 9:$ $$\begin{aligned} \sum_{k=1}^{9} F_k F_{10-k} &= 34 + 21 + 26 \\ &\quad {}+ 24 + 25 + 24 \\ &\quad {}+ 26 + 21 + 34 \\ &= 235. \end{aligned}$$