Coloring hypergraphs with no singleton intersections Let $H=(V,E)$ be a 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 $|e\cap f| > 1$?
 A: 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.  
A: There is an infinite example. Let ${\cal U}$ be a free ultrafilter on $\omega$, then $H=(\omega,{\cal U})$ is not $2$-colorable by Coloring non-principal ultrafilters on $\omega$ and no intersection of 2 edges is a singleton.
