You started out just fine by counting the following kinds of numbers:
___ ___ ___ ___ ___ ___ ___
6 6 6 1
6 6 6 2
6 6 6 3
6 6 6 4
6 6 6 5
And you’re right that you have to adjust for overcounting. Think of it this way, you’ve counted the numbers in the set $S_1$ of numbers $666????$, $S_2$ of numbers $?666???$, and so on. There are $10,000$ in each set. (Let’s assume ID numbers can begin with the digit $0$.)
Some numbers, however, are in more than one set. There are two kinds of these numbers: the numbers with $4$ or more consecutive sixes and the numbers with two separated groups of consecutive sixes. Let’s use X to mean a non-6 digit and count, beginning with those that have $4$, but not $5$, consecutive sixes.
Numbers of the form $6666X??$ are in $S_1$ and $S_2$.
Numbers of the form $X6666X?$ are in $S_2$ and $S_3$.
Numbers of the form $?X6666X$ are in $S_3$ and $S_4$.
Numbers of the form $??X6666$ are in $S_4$ and $S_5$.
There are 900+810+810+900 of these, and they were counted twice, so we need to subtract $3,420$ from your initial count of $50,000$. Now those with $5$, but not $6$ consecutive sixes. (We haven't considered them in the previous calculation.)
Numbers like $66666X?$, $X66666X$, and $?X66666$. Each was counted three times, and there are $90+81+90=261$ of them. So subtract another $522$.
Next, numbers like $X666666$ and $666666X$ were counted four times, and there are $9+9$ of those, so subtract another $3\cdot18$, and $6666666$ was counted five times, so subtract $4$.
Two separated $666$s occur in numbers like $666X666$ and were counted twice, so subtract $9$.
Summarizing, there are $50,000-3,420-522-54-4-9=45,591$ numbers containing $666$, leaving $9,954,009$ 7-digit numbers (possibly starting with one or more zeros) with no string of three sixes.
The mathematical principle (search and you'll find a general way to solve these problems) is called "inclusion-exclusion." Your initial approach is just what you need to get started with it.
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
Having difficulty with Inclusion-Exclusion for this problem, I instead found an approach using the distribution of non-distinct objects.
Consider these four cases:
In the first case our seven numbers are separated into four groups, and these groups separate the remaining 49 numbers into five groups, three of which will have at least one number: $$\small[x_1\text{ numbers}][g_1][x_2+1\text{ numbers}][g_2][x_3+1\text{ numbers}][g_3][x_4+1\text{ numbers}][g_4][x_5\text{ numbers}]$$ Here $x_1+x_2+x_3+x_4+x_5=46; x_i\ge0$ and this gives $\binom{50}{4}$ ways to choose the sizes of the five groups of numbers not in our subset. There are also $\binom43$ ways to chose which groups from our subset contain two consecutive numbers. This gives us a total of $\binom{50}{4}\binom43$ ways to choose a seven-element subset containing three pairs of consecutive numbers.
Similarly, in the second case we find $\binom{50}{5}\binom52$ ways to choose a seven-element subset containing two pairs of consecutive numbers.
In the third and fourth cases, we find $\binom{50}{6}\binom61$ and $\binom{50}{7}\binom70$ ways to choose our subset.
The total number of ways to choose a seven-element subset no containing three consecutive numbers is:$$\binom{50}{4}\binom43+\binom{50}{5}\binom52+\binom{50}{6}\binom61+\binom{50}{7}\binom70\\[5ex]203300\times 4+2118760\times 10+15890700\times 6+99884400\\[2ex]=217,337,400$$