Let $R$ be a commutative ring with 1, and let $h: R \to R$ be a ring automorphism of $R$. Set $S:= \{x\in R: h(x) = x\}$. That is, $S$ is the set of all elements of $R$ fixed by the automorphism $H$.
(a) Show that $S$ is a subring of $R$.
(b) Show that $1 \in S$.
(c) Now suppose that $h^2$ is the identity map and that $r \in R$. Prove that there is a monic polynomial $p \in S\lbrack x \rbrack$ of degree 2 with coefficients in $S$ such that $p(r) = 0$.
I have worked out a solution for (a) and (b), but I am unsure where to go for (c).
(a) Since $h$ is a ring homomorphism, $h(0) = 0$, so $0 \in S$ and $S \neq \varnothing$. Now, let $a,b \in S$. Now, \begin{align*} h(a-b) = h(a)-h(b) = a-b, \end{align*} so $a-b \in S$. Finally, let $x,y \in S$. Now, \begin{align*} h(xy) = h(x)h(y) = xy, \end{align*} so $xy\in S$, thus $S$ is a subring of $R$.
(b) Let $a \in S$, then, \begin{align*} h(a)h(1) = h(a\cdot1) = h(a), \end{align*} so $h(1)$ is the same as $1 \in R$. In other words, $h(1) = 1$, so $1 \in S$.
I would appreciate any feedback on these two parts. Part (c) has stumped me for some time, and I'm not quite sure were to go. I am not quite seeing how it is connected to the previous part either.
(b) is weird, since it follows from resp. is included in (a). For a subring, you also need to have $1 \in S$ by definition. Otherwise, you would need to call it a subrng, for example.
Your proof of $1 \in S$ is not correct. You get $1 \in S$ because, by definition, a ring/group/lattice/whatever homomorphism preserves the whole structure, in particular $h(1)=1$ for any ring homomorphism.
About the "$S \neq \emptyset$" in your proof, please have a look at my comment here. It also explains in more detail why (b) is included in (a), among other things.
For (c), notice that $r + h(r) \in S$ and $r \cdot h(r) \in S$. Hence, the polynomial $$p := X^2 - (r + h(r)) \cdot X + (r \cdot h(r))$$ is contained in $S[X]$. Now prove $p(r)=0$.
The motivation behind this polynomial comes from Galois theory, in case you wonder where it comes from. But of course Galois theory is not necessary to verify that it has the desired properties.