In your way of counting, if A sits in chair 3 you must have B in chair 4 so the choices are not independent. I don't see the advantage in the distinction between positive and negative conditions. The same problem could be posed that A wants to sit in $3,4, \text { or }5$ and B will not sit in $1,2, \text { or } 5$, which interchanges the positive and negative conditions. The key point is that whichever chair you select for B, there are the same number of options for A. If you seat A first, the number of options for B is not constant. You can divide it by cases, but seating B first avoids that.
Your argument is incorrect. To handle the restriction that $c_i \leq 1$, you need to use the Inclusion-Exclusion Principle because you need to exclude all cases in which one or more of the variables exceeds one. Doing so would be tricky since $k$ and $n$ are variables.
Here is an alternative approach.
Place $n - k$ empty chairs in a row. This creates $n - k + 1$ spaces in which we can place the $k$ people, $n - k - 1$ spaces between successive chairs and two at the ends of the row. The $n - k - 1$ spaces between successive chairs must each receive at least one person, so the answer is zero unless $k \geq n - k - 1$. The two spaces at the ends of the row need not receive a person. Therefore, if the number of people in the $i$th space from the left is $x_i$, the number of ways locations for the people may be distributed is the number of solutions of the equation
$$x_1 + x_2 + x_3 + \cdots + x_{n - k} + x_{n - k + 1} = k \tag{1}$$
in the integers subject to the constraints $x_2, x_2, \ldots, x_{n - k} \geq 1$ and $x_1, x_{n - k + 1} \geq 0$.
Let $x_1' = x_1 + 1$; let $x_{n - k + 1}' = x_{n - k + 1} + 1$; let $x_i' = x_i$ for $2 \leq i \leq n - k$. Then each $x_i'$, $1 \leq i \leq n - k + 1$, is a positive integer. Substituting $x_1' - 1$ for $x_1$, $x_{n - k + 1}' - 1$ for $x_{n - k + 1}$, and $x_i'$ for $x_i$, $2 \leq i \leq n - k$, in equation 1 yields
\begin{align*}
x_1' - 1 + x_2' + x_3' + \cdots + x_{n - k}' + x_{n - k + 1}' - 1 & = k\\
x_1' + x_2' + x_3' + \cdots + x_{n - k}' + x_{n - k + 1}' & = k + 2 \tag{2}
\end{align*}
A particular solution to equation 2 in the positive integers corresponds to the placement of $n - k$ addition signs in the $k + 1$ spaces between successive ones in a row of $k + 2$ ones. For instance, if $n = 12$ and $k = 7$, then $n - k = 5$ and $k + 2 = 9$, so
$$1 + 1 1 + 1 + 1 1 + 1 + 1 1$$
corresponds to the solution $x_1' = 1$, $x_2' = 2$, $x_3' = 1$, $x_4' = 2$, $x_5 = 1$, $x_6' = 2$ of equation 2 and the solution $x_1 = 0$, $x_2 = 2$, $x_3 = 1$, $x_4 = 2$, $x_5 = 1$, $x_6 = 1$ of equation 1. The number of solutions of equation 2 is the number of ways we can select $n - k$ of the $k + 1$ spaces between successive ones in a row of $k + 2$ ones, which is $$\binom{k + 1}{n - k}$$
The $k$ people can then be arranged in those locations in $k!$ ways. Therefore, the number of ways that $k$ people can be seated in a row of $n$ chairs so that no two empty chairs are adjacent is
$$\binom{k + 1}{n - k}k!$$
Best Answer
Let's start with person D who is the most restrictive. We will consider 2 cases:
D sits in chair 2 or 4
D sits in chair 3
If D sits in chair 2 or 4, next, we look at person A who cannot sit in chair 3. There are only 3 chairs available to A. Then 3 to B, 2 to C, and 1 to E. That is: $2\cdot 3\cdot 3\cdot 2\cdot 1 = 36$
If D sits in chair 3, then there are $4!=24$ ways to arrange the remaining people, as there are no more restrictions.
That's a total of 36+24=60.
Now, let's try with your method (we take the total number of permutations and subtract restricted cases).
As you found, there are 120 total permutations. There are $2\cdot 4\cdot 3\cdot 2\cdot 1=48$ permutations where D is in either chair 1 or 5. There are $1\cdot 4\cdot 3\cdot 2\cdot 1 = 24$ permutations where A is in chair 3.
The number of permutations with A in chair 3 OR D in chair 1 or 5 equals the number of permutations of A in chair 3 plus the number of permutations of D in chair 1 or 5 minus the permutations of both. So, there are $2\cdot 1\cdot 3\cdot 2\cdot 1 = 12$ permutations where both occur.
$120-(48+24-12) = 60$.