Without finiteness assumptions, the irreducible and the connected components of a scheme may behave in strange ways. More precisely, let us consider a scheme $X$ and the following properties:


(1) $X$ is the sum of its irreducible components;


(2) The irreducible and the connected components of $X$ coincide;


(3) The irreducible components of $X$ are pairwise disjoint.


It is clear that (1) implies (2), that (2) implies (3), and that if the set of irreducible components of $X$ is locally finite then all three statements are equivalent (see [EGA 0.2.1.6]). However, (2) does not necessarily imply (1) in general: An affine counterexample is given by the spectrum of the product of infinitely many fields (which is non-discrete and totally disconnected). So, out of pure curiosity we may ask the following:



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> Is there an (affine) scheme fulfilling (3) but not (2)?
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One can note that this is equivalent to the following:



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> Is there a nonempty, reducible, connected (affine) scheme whose irreducible components are pairwise disjoint?
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(One can also note that for topological spaces that are not necessarily underlying spaces of schemes it is easy to find an example that fulfils (3) but not (2) - every connected, separated space with at least two points does so.)



A reducible connected scheme with pairwise disjoint irreducible components
In the answers to [Non-integral scheme having integral local rings](https://mathoverflow.net/questions/7477/non-integral-scheme-having-integral-local-rings) , and in particular in t3suji's beautiful geometric construction, it is shown that there is a (non-empty) affine scheme $X=\mathrm{Spec}(A)$ whose underlying topological space is connected, such that all local rings of $X$ are integral domains and such that $X$ is however not integral. This scheme fulfills the conditions you are requiring: It is reduced and not integral, hence reducible. Assume two irreducible components $Y\_1$ and $Y\_2$ of $X$ met in a point $x\in X$. The point $x$ corresponds to a prime ideal $\mathfrak{q} \subset A$, and the irreducible components $Y\_1$ and $Y\_2$ are of the form $Y\_1=V(\mathfrak{p}\_1)$ and $Y\_2=V(\mathfrak{p}\_2)$ for minimal prime ideals $\mathfrak{p}\_1$ and $\mathfrak{p}\_2$ of $A$. As $x\in Y\_1,Y\_2$, we would have $\mathfrak{p}\_1, \mathfrak{p}\_2 \subset \mathfrak{q}$. 


Then $\mathfrak{p}\_1$ and $\mathfrak{p}\_2$ would correspond to distinct minimal prime ideals of $A\_\mathfrak{q}=\mathcal{O}\_{X,x}$, which is impossible, as $\mathcal{O}\_{X,x}$ is by construction a domain, so that its only minimal prime ideal is $\{0\}$.


