I've partly solved the following exercise of Vakil's FOAG, but I am not sure I got the last part right. Could some take a look?
5.5.C. EXERCISE (ASSUMING (A)). Show that if $A$ is reduced, Spec $A$ has no embedded points. Hints: (i) first deal with the case where $A$ is integral, i.e., where
Spec $A$ is irreducible. (ii) Then deal with the general case. If $f$ is a nonzero function on a reduced affine scheme, show that $\operatorname{Supp}f = \overline{D(f)}$: the support is the closure of the locus where $f$ doesn’t vanish. Show that $\overline{D(f)}$ is the union of the irreducible components meeting $D(f)$, using (i).
My solution goes as follows:
(i) Since $A$ is integral, an embedded point is given by a non-unit $f$ such that $\operatorname{Supp}f \not \supset V(f)$. This is because Supp $f\supset D(f)$, so that if it also contained $V(f)$, it would contain everything. On the other hand, if $\operatorname{Supp} f\not \supset V(f)$, then Supp $f$ can't be an irreducible component of Spec $A$ and must be the closure of a union of embedded points.
But Supp $f\not \supset V(f)$ means that $\exists \mathfrak p \ni f$ with $f_\mathfrak p =0 \iff f\cdot a =0$ for some $a\not \in \mathfrak p$. But this is not possible because $A$ was assumed to be a domain.
(ii) In this post is a proof of $\operatorname{Supp}f = \overline{D(f)}$ with the above conditions. I claim that, since there are only finitely many minimal primes (noetherianness is a standing assumption in this chapter), we can cover Spec $A$ by finitely many clopen $D(g_i)$, each one of which corresponding to the closure of a minimal prime. It follows that $(g_i) = A$, and we can write $$\overline{D(f)}=\overline{D(fA)}=\overline{D(f(g_i))}=\overline{D((fg_i))}=\bigcup\overline{D(fg_i)}$$
Now $\overline{D(fg_i)}$ is the $\operatorname{Supp} f \cap \operatorname{Spec} A_{g_i}$, which is the spectrum of an integral domain. By (i) this cannot correspond to a proper irreducible subset of $\operatorname{Spec} A_{g_i}$, so (ii) follows.
The claim can be proven by induction, and I show the case where $A$ has three minimal primes $\mathfrak p, \mathfrak q, \xi$. By symmetry, it suffices to show that we can pick an element in $\mathfrak p$ that is not in $\mathfrak q, \xi$. So pick $p_1, p_2 \in \mathfrak p$ such that $p_1 \not \in \mathfrak q$ and $p_2 \not \in \xi$. If either one of the $p_i$ is not in the other two primes, we are done. Otherwise $p_1+p_2$ does the job.
Best Answer
Let me give an algebraic answer to your question: to me, an embedded point of $\text{Spec}(A)$ is a nonminimal associated prime (your definition does not quite match up with this: points of $\text{Spec}(A)$ do not correspond to elements of $A$).
Here are two reasons why a reduced ring $A$ cannot have embedded primes:
1) Any embedded prime is the annihilator of a nilpotent element. Indeed, every associated prime $p$ is of the form $\text{ann}(x)$, for some $x \in A$. If $p = \text{ann}(x)$ is embedded, then $px = 0$ is contained in every minimal prime, but $p$ is not, so $x$ is in every minimal prime, hence $x$ is nilpotent (this even shows $x^2 = 0$).
2) The associated primes of any ideal $I$ are the radicals of the primary ideals appearing in a minimal primary decomposition of $I$. Since $A$ is reduced, $0 = \sqrt{0} = p_1 \cap \ldots \cap p_n$, where $p_i$ are the minimal primes of $A$. But then every associated prime of $A$ is one of the $p_i$, hence minimal.