The proof given in David Burton is as follows
Suppose that $a^n\equiv a$ mod($n$), for every integer $a$, but $k^2\,|\,n$ for some integer $k>1$.
If we let $a=k$ then $k^n\equiv k$ mod($n$). Because $k^2\,|\,n$, the last congruence holds modulo $k^2$; that is $k\equiv k^n ≡ 0$ mod ($k^2$), therefore $k^2\,|\,k$ which is impossible. Thus $n$ must be square free.
My question is:
In the assumption step, instead of every integer $a$, shouldn't it be every integer $a$ such that gcd$(a,n)=1$?
As gcd of $a$ and $n$ is always $1$ how can we assume $a=k$, because in that case gcd$(a,n)$ wouldn't be $1$?
Not sure that "absolute pseudo-prime" has a well established definition...I have seen the phrase used for several, related, notions. As to Carmichael number, I'd agree with your requirement. Specifically, I'd say that $n$ was a Carmichael number if: $$(a,n)=1 \implies a^{n-1}=1\mod(n)$$
Using that definition: we note that, if $p$ is a prime such that $p^2|n$ then $p|\varphi(n)$ whence $\exists g\in \mathbb Z_n^*$ such that $g\neq 1$ but $g^p=1$ But since $p$ clearly does not divide $n-1$ we get a contradiction.