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.
Best Answer
I will solve the problem in two ways.
Method 1: We use the Inclusion-Exclusion Principle.
There are $n!$ permutations of the elements of the set $\{1, 2, 3, \ldots, n\}$. From these, we must exclude those permutations in which at least two of the numbers $1, 2, 3$ are consecutive.
Since there are only three numbers in the subset, there can be at most two pairs of consecutive numbers from the subset $\{1, 2, 3\}$ in the permutation, with two pairs occurring when all three of those numbers appear consecutively in the permutation.
A pair of consecutive numbers from the subset $\{1, 2, 3\}$: There are $\binom{3}{2}$ ways to select which two of those three numbers are consecutive. We now have $n - 1$ objects to permute, a block of two consecutive numbers and the other $n - 2$ numbers in the set $\{1, 2, 3, \ldots, n\}$. The $n - 1$ objects can be permuted in $(n - 1)!$ ways. The numbers in the block can be permuted in $2!$ ways. Hence, there are $$\binom{3}{2}(n - 1)!2!$$ such permutations.
However, if we subtract that amount, we will have subtracted each permutation in which there are two pairs of consecutive numbers drawn from the subset $\{1, 2, 3\}$ twice, once when we designate the first two of the three consecutive numbers as the pair of consecutive numbers and once when we designate the last two of the three consecutive numbers as the pair of consecutive numbers. We only want to subtract such permutations once, so we must add them to the total.
Two pairs of consecutive numbers from the subset $\{1, 2, 3\}$: For this to occur, the three numbers in the subset must appear in consecutive positions. Consequently, we have $n - 2$ objects to permute, the block consisting of the numbers $1, 2, 3$ and the other $n - 3$ numbers in the set $\{1, 2, 3, \ldots, n\}$. The $n - 2$ objects can be permuted in $(n - 2)!$ ways. The numbers in the block can be permuted in $3!$ ways. Hence, there are $$\binom{3}{3}(n - 2)!3!$$ such permutations.
By the Inclusion-Exclusion Principle, the number of admissible permutations is $$n! - \binom{3}{2}(n - 1)!2! + \binom{3}{3}(n - 2)!3!$$
Method 2: We arrange the $n - 3$ numbers in the subset $\{4, 5, 6, \ldots, n\}$, then place the numbers $1, 2, 3$ in the spaces this creates.
There are $(n - 3)!$ ways to arrange the $n - 3$ numbers in the subset $\{4, 5, 6, \ldots, n\}$. This creates $n - 2$ spaces in which to place the numbers $1, 2, 3$, including $n - 4$ spaces between successive numbers in the permutation of the numbers in the subset $\{4, 5, 6, \ldots, n\}$ and two at the ends of the row. To ensure that no two of the three numbers $1, 2, 3$ are consecutive, we must choose three of these $n - 2$ spaces in which to place the numbers $1, 2, 3$, which can be done in $\binom{n - 2}{3}$ ways. The three numbers $1, 2, 3$ can be arranged in the selected spaces in $3!$ ways. Hence, the number of admissible permutations is $$(n - 3)!\binom{n - 2}{3}3!$$
We show that the above methods yield equivalent results.
\begin{align*} n! - \binom{3}{2}(n - 1)!2! + \binom{3}{3}(n - 2)!3! & = n! - 3 \cdot 2 \cdot (n - 1)! + 1 \cdot 6 \cdot (n - 2)!\\ & = n(n - 1)(n - 2)! - 6(n - 1)(n - 2)! + 6(n - 2)!\\ & = (n - 2)![n(n - 1) - 6(n - 1) + 6]\\ & = (n - 2)!(n^2 - n - 6n + 6 + 6)\\ & = (n - 2)!(n^2 - 7n + 12)\\ & = (n - 2)!(n - 3)(n - 4) \end{align*}
\begin{align*} (n - 3)!\binom{n - 2}{3}3! & = (n - 3)! \cdot \frac{(n - 2)!}{3!(n - 5)!} \cdot 3!\\ & = (n - 3)!(n - 2)(n - 3)(n - 4)\\ & = (n - 2)!(n - 3)(n - 4) \end{align*} so the results we obtained are equivalent.
Let's compare your result with the above result.
In order for no two of the numbers $1, 2, 3$ to be consecutive in a permutation of the set $\{1, 2, 3, \ldots, n\}$, $n$ must be at least $5$.
For $n = 5$ and $n = 6$, our answers agree.
However, if $n = 7$, my formula yields $$(n - 2)!(n - 3)(n - 4) = 5! \cdot 4 \cdot 3 = 1440$$ Using the table of Stirling numbers of the second kind, we see that your result gives \begin{align*} 3!(n - 3)![S(n - 3, 2) + S(n - 3, 3) + S(n - 3, 4)] & = 3!4![S(4, 2) + S(4, 3) + S(4, 4)]\\ & = 3!4!(7 + 6 + 1)\\ & = 2016 \end{align*} so your method overcounts.
I suspect the error lies in permuting elements which are in different blocks. Once you separate the numbers $4, 5, 6, \ldots, n$, they can only be permuted within each block.