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Define a selfish set to be a set which has its own cardinality (number of elements) as an element. Find, with proof, the number of subsets of $1, 2, ldots, n$ which are minimal selfish sets, that is, selfish sets none of whose proper subsets is selfish.

My Attempt:
Assume $textbfA$ to be a selfish set. If the cardinality of $textbfA$ is $c$, then can $textbfA$ contain $1,2,3....c-1$. Definitely answer is no. because if it contains $k<c$ then deleting $c-k$ elements except $k$ from $textbfA$ gives a subset of k elements contradicting the fact that $textbfA$ is minimal selfish.
Thus $textbfA$ must contain elements greater than or equal to $c$. But how do I find the minimal selfish sets with order $c$?

Your argument is correct.

Lets see if recursion helps.

Let $[n]$ denote the set $1,2,ldots,n$, and let $f_n$ denote the
number of minimal selfish subsets of $[n]$. Then the number of
minimal selfish subsets of $[n]$ not containing $n$ is equal to
$f_n-1$. On the other hand, for any minimal selfish subset of $[n]$
containing $n$, by subtracting 1 from each element, and then taking
away the element $n-1$ from the set, we obtain a minimal selfish
subset of $[n-2]$ (since $1$ and $n$ cannot both occur in a selfish
set). Conversely, any minimal selfish subset of $[n-2]$ gives rise to
a minimal selfish subset of $[n]$ containing $n$ by the inverse
procedure. Hence the number of minimal selfish subsets of $[n]$
containing $n$ is $f_n-2$. Thus we obtain $f_n=f_n-1+f_n-2$.
Since $f_1=f_2=1$, we have $f_n=F_n$, where $F_n$ denotes the $n$th
term of the Fibonacci sequence.

Your logic so far is fine. So what you know is that, since $c$ is in the set, then the other $c-1$ elements must all be at least $c+1$. There are $binomn-cc-1$ ways to choose them.

Summing over these gives you the total count. It turns out that this gives you the $n^th$ Fibonacci number, which you can prove by induction (hint: use Pascal’s identity).