Yep this is correct!
Consider the following algebraic characterization of $0$-dimensional rings.
A ring $R$ is $0$-dimensional iff for every $x \in R$ there exists an $n \in \mathbb{N}$ such that $x^{n+1} R = x^n R$
Proof I think you already know the if direction. Conversely, suppose for the sake of contradiction that $x^{n+1} y \not= x^n$ for any $y \in R, n \in \mathbb{N}$. Then in particular the multiplicative set generated by the elements $x, 1-xy$ does not contain $0$. So there's a prime $\mathfrak{p}$ disjoint from this multiplicative set. That would firstly imply $x \notin \mathfrak{p}$, but since the ring is $0$-dimensional by assumption, $\mathfrak{p}$ is maximal and then $x$ is a unit in $R/\mathfrak{p}$, hence $ax - 1 \in \mathfrak{p}$ for some $a \in R$, also a contradiction. $\square$
In your case $a^p = a$, so in fact $a^2R = aR$.
Also note that any $0$-dimensional ring is Hausdorff with respect to the Zariski topology. This is a special case of the fact that
The minimal prime spectrum of a ring is always Hausdorff with respect to the (topology induced by the) Zariski topology.
Proof Indeed, if $\mathfrak{p}, \mathfrak{q}$ are two distinct minimal primes of a ring, then find an element $f \in \mathfrak{p} \setminus \mathfrak{q}$. Since $f \in \mathfrak{p}$, the image of $f \in R_\mathfrak{p}$ is in the nilradical of $R_\mathfrak{p}$, hence there exists $s \notin \mathfrak{p}$ such that $sf^n = 0$ for some $n$. In particular, $\mathfrak{q} \in D(s), \mathfrak{p} \in D(f)$, and $D(f) \cap D(s) = \emptyset$, so we've separated $\mathfrak{p}, \mathfrak{q}$ $\square$.
You can also show that the minimal prime spectrum of a ring is totally disconnected with respect to the Zariski topology. So $0$-dimensional rings are quasi-compact, Hausdorff, and totally disconnected (so yes they are $0$-dimensional as topological spaces too).
As you say, Stone duality assures us that the spectrum of any $0$-dimensional ring can be realized as the spectrum of a Boolean algebra. In particular, we can put a Boolean algebra structure on the idempotents of any ring by defining addition as the map sending $(e,f) \to e + f - 2ef$ (and multiplication inherited from the ring). This is isomorphic to the canonical algebra structure on the clopens of the Zariski topology. Stone duality is telling us that the idempotent algebra associated to a $0$-dimensional ring has the same spectrum as the original ring.
Yes, a closed set in the Zariski topology is a set of the form $V(I) = \{\mathfrak{p}\in\rm{Spec}(R) \mid I\subset \mathfrak{p}\}$. If you fix the ideal $I$, then there is only one closed set in the topology! That clearly isn't right.
To see that closed points give maximal ideals, just notice that all ideals are contained in maximal ideals. Therefore, all closed sets contain at least one maximal ideal. So if a single point is a closed set, it has to be a maximal ideal.
Best Answer
First of all, I would like to recall the construction of the Zariski topology: Closed sets in $\operatorname{Spec}A$ are given by $ Z(\mathfrak{a}):=\{\mathfrak{p}\in\operatorname{Spec}A\mid\mathfrak{a}\subseteq\mathfrak{p}\} $ for ideals $\mathfrak{a}\subseteq A$. This is basically the only fact we need to prove the separation property.
Assume that $\mathfrak{p}$, $\mathfrak{q}\in\operatorname{Spec}A$ are distinct prime ideals of $A$ that are both not maximal. We have to be careful because it could be that one ideal contains the other; however, it cannot be that both $\mathfrak{p}\subset\mathfrak{q}$ and $\mathfrak{q}\subset\mathfrak{p}$. Without loss of generality we can assume $\mathfrak{p}\not\subset\mathfrak{q}$. But then by definition $\mathfrak{q}\notin Z(\mathfrak{p})$ and hence $Z(\mathfrak{p})$ is a closed set that contains $\mathfrak{p}$ but not $\mathfrak{q}$. Now we can construct the desired neighbourhood $N:=\operatorname{Spec}A\setminus Z(\mathfrak{p})$; by construction $\mathfrak{q}\in N$ and $\mathfrak{p}\notin N$.