Show that $f(x) = x^p -x -1 \in \Bbb{F}_p[x]$ is irreducible over $\Bbb{F}_p$ for every $p$.

I like your idea very much, but I didn't understand how you managed to ignore the $\alpha^2$ term from your constant term.

I have written up an answer to this question with a similar idea, where I look at the next to highest degree term of $g(x)$ (=the term of degree $k-1$). In your degree two example, this would be the linear term. Its coefficient is $2\alpha+(s_i+s_j).$ As $s_i,s_j$ are in the prime field, and $2$ is invertible, we can, as in your argument, conclude that $\alpha\in\mathbb{F}_p$, which is a contradiction.

By studying the lowest degree term, you get a lot of clutter from powers of $\alpha$. The degree $k-1$ term is IMHO easier to manage.

a) and b) are fine. I haven't checked c) but you could also note that if the polynomial splits into $h \cdot g$ with $h,g \in \mathbb{F}_p[x]$ then, say $$ h(x) = \prod_{i \in S} (x - (\alpha + i)) \in \mathbb{F}_p[x] $$

with $S$ proper subset of $\{0,1,\ldots,p-1\}$. Therefore $$ \sum_{i \in S} (\alpha + i) \in \mathbb{F}_p $$ hence $|S| \alpha \in \mathbb{F}_p$ but since $0 < |S| < p$ this implies that $\alpha \in \mathbb{F}_p$, but then $$ f(x) = \prod_{i} (x - i) = x^p - x $$ which is not true.

(This is an alternate approach that doesn't use the outlined argument.)

In general, if $h(x)|x^{p^n}-x$ in $\mathbb F_p[x]$ then $h(x)$ is the product of distinct prime polynomials, each of degree equal to a factor of $n$.

If you know this, a fun way to prove this theorem (but skipping the outlined steps in the question) is to prove that $x^p-x-1$ is a divisor of $x^{p^p}-x$. Since $x^p-x-1$ has no roots in $\mathbb F_p$, this would mean that it has to be a prime polynomial.

You show that it is a factor by noting that:

$$x^{p^p}-x = \sum_{k=0}^{p-1} \left(x^{p^{k+1}}-x^{p^k}\right) = \sum_{k=0}^{p-1} (x^p-x)^{p^k} = \\\sum_{k=0}^{p-1}((x^p-x-1)+1)^{p^k} \equiv 0\pmod{x^p-x-1}$$