Let's apply your strategy of using the Inclusion-Exclusion Principle.
There are $\binom{10}{4}$ four-element subsets of a ten-element set. From these, we must exclude those subsets containing at least two consecutive numbers.
A string of at least two consecutive numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ can begin with any of the nine numbers less than $10$. Once we select a pair of consecutive numbers beginning with one of these nine numbers, we must select two additional elements of the ten-element set from the eight elements that are not in the pair of consecutive numbers, which we can do in $\binom{8}{2}$ ways. Thus, it appears we have $9 \cdot \binom{8}{2}$ sets containing at least two consecutive numbers. However, we have overcounted since we have counted sets containing two disjoint pairs twice. There are $\binom{8}{2}$ such sets since we must exclude from the $\binom{9}{2}$ ways of selecting two of the nine numbers less than $10$ as starting points for the two pairs of consecutive integers the $\binom{8}{1}$ ways of selecting consecutive starting points that would prevent them from being disjoint. Note that it is a consequence of Pascal's Identity that $\binom{9}{2} - \binom{8}{1} = \binom{8}{2}$. Hence, there are
$$9 \cdot \binom{8}{2} - \left[\binom{9}{2} - \binom{8}{1}\right] = 9 \cdot \binom{8}{2} - \binom{8}{2} = 8 \cdot \binom{8}{2}$$
four-element subsets containing at least two consecutive numbers.
A string of at least three consecutive numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ can begin with any of the eight numbers less than $9$. Once we select a triple of consecutive numbers, we can select the remaining element in the subset from one of the seven elements in the set that are not in the triple, giving
$$8 \cdot \binom{7}{1}$$
four-element subsets that contain at least three consecutive numbers.
A string of four consecutive numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ can begin with any of the seven numbers less than $8$.
Thus, by the Inclusion-Exclusion Principle, the number of four-element subsets of the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ that do not contain consecutive numbers is
$$\binom{10}{4} - 8 \cdot \binom{8}{2} + 8 \cdot \binom{7}{1} - 7 = 210 - 224 + 56 - 7 = 35$$
Addendum: A more efficient approach is to arrange six blue and four green balls in a row so that no two of the green balls are consecutive, then number the balls from left to right. The numbers on the green balls are the desired subset of four numbers, no two of which are consecutive, selected from the set $[10] = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.
Line up six blue balls in a row. This creates seven spaces, five between successive blue balls and two at the ends of the row.
$$\square b \square b \square b \square b \square b \square b \square$$
To ensure that no two of the green balls are consecutive, we select four of these seven spaces in which to place a green ball, which can be done in
$$\binom{7}{4} = 35$$
ways. We now number the balls from left to right. The numbers on the green balls are the desired subset of four numbers selected from the set $[10]$, no two of which are consecutive.
An $n$-word is a word of length $n$. Call a set with no consecutives good.
There are two types of good set (i) the ones that do not contain $1$, and (ii) the ones that contain $1$.
There are $a_{n-1}$ good subsets of $\{1,2,\dots,n\}$ of Type $1$. There are $a_{n-2}$ good subsets of $\{1,2,\dots,n\}$ of Type (ii), since if our set contains $1$, then $2$ is forbidden.
Remark: The author has chosen to phrase things in terms of characteristic functions instead of subsets. That makes no difference, characteristic functions and subsets are essentially the same.
Best Answer
An example with a specific $n$ is illustrative.
Suppose $n = 7$. Then an ordered non-consecutive subset of $\{1, 2, 3, 4, 5, 6, 7\}$ that contains $7$ might be
$$(1, 4, 7).$$
Notice that the fact that the last element is $7$ means that $6$ is excluded from the possible choices for the other elements, and that one must choose among the set $\{1, 2, 3, 4, 5\}$ for the other elements; in this case, $(1, 4)$ was chosen. Consequently, the number of ordered non-consecutive subsets of $\{1, 2, 3, 4, 5, 6, 7\}$ that contain $7$ is equal to the number of ordered non-consecutive subsets of $\{1, 2, 3, 4, 5\}$.
On the other hand, if an ordered non-consecutive subset of $\{1, 2, 3, 4, 5, 6, 7\}$ does not contain $7$, then obviously its elements are chosen among $\{1, 2, 3, 4, 5, 6\}$.
Thus the total number of non-consecutive subsets of $\{1, 2, 3, 4, 5, 6, 7\}$ is equal to the sum of the number of non-consecutive subsets of $\{1, 2, 3, 4, 5\}$ and of $\{1, 2, 3, 4, 5, 6\}$.
As there was nothing special about our choice $n = 7$, we see how the argument generalizes to any $n \ge 3$.
However, we must also take care to define what happens when $n \le 2$. In the $n = 2$ case, we can count $(1)$ or $(2)$, so there are at least two non-consecutive subsets of $\{1, 2\}$ in the sense that there lacks two elements whose difference is $1$. But should we count the empty set? Does it make sense to say the empty set contains no consecutive elements? The book's answer seems to suggest so by simply claiming $f(2) = 3$.