Let $H=(V,E)$ be a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph). If $\kappa>0$ is a cardinal, we say the hypergraph $H$ is $\kappa$-*chromatic* if there is a function $c:V\to\kappa$ such that for all $e\in E$ the restriction $c|\_e$ is not constant (that is, the vertices of every edge are colored with at least $2$ colors).


So far, all the hypergraphs I have come across that are $\kappa$-chromatic for some cardinal $\kappa>2$, but not $2$-chromatic have $2$ distinct edges such that their intersection is a singleton.


**Question**. Does there exist a hypergraph $H=(V,E)$ that is $\kappa$-chromatic for some cardinal $\kappa>2$, but not $2$-chromatic, such that for all $e,f\in E$ we have $e\cap f = \emptyset$ or $|e\cap f| > 1$?



Coloring hypergraphs with no singleton intersections
To answer Jon Noel's question in the comments, there is no such example for finite hypergraphs. 


**Claim.** Let $H=(V,E)$ be a finite hypergraph such that $|e| > 1$ for all $e \in E$ and for all distinct $e\_1, e\_2 \in E$, $|e\_1 \cap e\_2| \neq 1$. Then $H$ is $2$-chromatic. 


*Proof*. We proceed by induction on $|E|$. The base case is clear. Now arbitrarily choose $e \in E$ and let $G=(V, E \setminus e)$. By induction, $G$ has a $2$-coloring $c$. If two vertices in $e$ receive distinct colors from $c$, then $c$ is a $2$-colouring of $H$ and we are done. So we may assume that all vertices in $e$ are red. We now choose a vertex $x \in e$ and recolor $x$ blue. This is a valid $2$-colouring of $H$ unless there is an edge $f \in E$ such that $x \in f$ and all other vertices in $f$ are blue. But this implies that $|e \cap f|=1$, which is a contradiction. 


