I can show that if $X$ is a scheme such that all local rings $\mathcal{O}\_{X,x}$ are integral and such that the underlying topological space is connected and Noetherian, then $X$ is itself integral. 


This doesn't seem to work without the "Noetherian" condition. But can anyone think about a nice counterexample to illustrate this? So I am looking for a non-integral scheme - with connected underlying topological space - having integral local rings. 



Non-integral scheme having integral local rings
Let me try to give a counterexample. (I don't know whether it is 'nice'). First, let us rewrite your properties for an affine scheme $X=Spec(A)$.


Connectedness for $A$ means $A$ has no nontrivial idempotents;


Integrality for $A$ is the usual one ($A$ is a domain);


Local integrality means that whenever $fg=0$ in $A$, every point of $X$ has a neighborhood
where either $f$ or $g$ vanishes.


Let us construct a connected locally integral ring that is not integral.


Roughly speaking, the construction is as follows: let $X\_0$ be the cross (the union of coordinate axes) on the affine plane. Then let $X\_1$ be the (reduced) full preimage of $X\_0$ on the blow-up of the plane ($X\_1$ has three rational components forming a chain). Then blow up the resulting surface at the two singularities of $X\_1$, and let $X\_2$ be the reduced preimage of $X\_1$
(which has five rational components), etc. Take $X$ to be the inverse limit.


The only problem with this construction is that blow-ups glue in a projective line, so $X\_1$ is not affine. Let us correct this by gluing in an affine line instead (so our scheme will be an open subset in what was described above). 


Here's an algebraic description:


For every $k\ge 0$, let $A\_k$ be the following ring: its elements are collections of
polynomials $p\_i\in{\mathbb C}[x]$ where $i=0,\dots,2^k$ such that $p\_i(1)=p\_{i+1}(0)$.
Set $X\_k=Spec(A\_k)$. $X$ is a union of $2^k+1$ affine lines that meet transversally in a chain. (It may be better to index polynomials by $i/2^k$, but the notation gets confusing.)


Define a morphism $A\_k\to A\_{k+1}$ by
$$(p\_0,\dots,p\_{2^k})\mapsto(p\_0,p\_0(1),p\_1,p\_1(1),\dots,p\_{2^k})$$
(every other polynomial is constant). This identifies $A\_k$ with a subring of $A\_{k+1}$.
Let $A$ be the direct limit of $A\_k$ (basically, their union). Set $X=Spec(A)$. For every
$k$, we have a natural embedding $A\_k\to A$, that is, a map $X\to X\_k$.


Each $A\_k$ is connected but not integral; this implies that $A$ is connected but not integral. It remains to show that $A$ is locally integral.


Take $f,g\in A$ with $fg=0$ and $x\in X$. Let us construct a neighborhood of $x$ on which one of $f$ and $g$ vanishes. Choose $k$ such that $f,g\in A\_{k-1}$ (note the $k-1$ index).
Let $y$ be the image of $x$ on $X\_k$. It suffices to prove that $y$ has a neighborhood on
which either $f$ or $g$ (viewed as functions on $X\_k$) vanishes.


If $y$ is a smooth point of $X\_k$ (that is, it lies on only one of the $2^k+1$ lines), this is obvious. We can therefore assume that $y$ is one of the $2^k$ singular points, so two components of $X\_k$ pass through $y$. However, on one of these two components (the one with odd index), both $f$ and $g$ are constant, since they are pullbacks of functions on $X\_{k-1}$. Since $fg=0$ everywhere, either $f$ or $g$ (say, $f$) vanishes on the other component.
This implies that $f$ vanishes on both components, as required. 


