Let $X$ be a connected scheme. Then $X$ irreducible iff $\forall x, \text{ Spec}(\mathcal{O}_{X,x})$ is I could use some help for the second part of exercise I-34 from Eisenbud and Harris' Geometry of Schemes:

Let $X$ be a connected scheme. Show that $X$ is irreducible if and only if 
  for all $x \in X$, the stalk local ring has a unique minimal prime ideal.

I guess it must be related to the first part of the exercise, that states:

An arbitrary scheme is irreducible iff every open affine subset is irreducible. 


My ideas
I know that an affine scheme $\text{Spec }R$ is irreducible precisely when $R$ has a unique minimal prime ideal, so that leaves us with the to be proven statement:
$$
X \text{ is irreducible}
\quad\iff \quad
\forall x \in X, \text{ Spec}(\mathcal{O}_{X,x}) \text{ is irreducible}.
$$
but the RHS seems even harder to prove:  $\text{ Spec}(\mathcal{O}_{X,x})$ doesn't really seem to be "accessible" in some way. For example, this result cannot be applied since $\text{ Spec}(\mathcal{O}_{X,x})$ cannot be identified with affine patches as one would like to do. Do you have ideas?

This is false.  Indeed, there exists a ring $A$ which has no nontrivial idempotent, which is not a domain, and such that the localization of $A$ at any prime ideal is a domain.  See http://stacks.math.columbia.edu/tag/0568 for details of the construction.  Note that such a ring automatically is reduced, since all of its localizations at prime ideals are reduced.
Since $A$ has no nontrivial idempotent elements, $\operatorname{Spec} A$ is connected, and since every localization of $A$ at a prime ideal is a domain, every stalk has a unique minimal prime.  But $\operatorname{Spec}A$ is reducible, since any nonzero $f,g\in A$ with $fg=0$ give proper closed subsets whose union is all of $\operatorname{Spec} A$ (the vanishing sets of $f$ and $g$ are proper subsets since $f$ and $g$ are not nilpotent).

On the other hand, here are some positive results.  First, the forward direction is always true.  Indeed, if $X$ is irreducible and $x\in X$, let $U=\operatorname{Spec} A$ be an affine open subset containing $x$.  Then $U$ is irreducible, so $A$ has a unique minimal prime.  The same is then true of any (nonzero) localization of $A$, in particular the local ring $\mathcal{O}_{X,x}$.
Second, the reverse direction is true if $X$ is Noetherian.  Indeed, in that case $X$ can be written as a finite union of irreducible components.  If $X$ is not irreducible, it has more than one irreducible component, and if $X$ is connected, there must be two distinct irreducible components $A$ and $B$ of $X$ which intersect (otherwise, each irreducible component would be clopen).  Let $x\in A\cap B$.  Then the generic point of $A$ and the generic point of $B$ give distinct minimal primes in the local ring $\mathcal{O}_{X,x}$.
