If $X\neq \varnothing$ is a set we say that ${\frak P} \subseteq {\cal P}(X)$ is a partition of $X$ if
$\bigcup{\frak P} = X$, and
$P\neq Q \in {\frak P} \implies P\cap Q = \varnothing$.
Let $H = (V,E)$ be a hypergraph with $V \neq \varnothing$ and $\bigcup E = V$. A partition ${\frak P}$ of $V$ is said to be splitting if for all $e\in E$ and $P \in {\frak P}$ we have $|e \cap P| = 1$, and we call such a hypergraph admitting a splitting partition partite. It is easy to see that the edges of a partite hypergraph all have the same cardinality - and every splitting partition also has that cardinality.
If $E_1 \subseteq E$ we call $H|_{E_1}:=(\bigcup E_1, E_1)$ the subhypergraph induced by $E_1$.
Question. If $H=(V,E)$ is a hypergraph with $\bigcup E = V$, is there $E_1\subseteq E$ such that $H|_{E_1}$ is partite, but whenever we have $E_2\subseteq E$ with $E_1\subseteq E_2$ and $E_1\neq E_2$, then $H|_{E_2}$ is no longer partite?
Note. A straightforward application of Zorn's Lemma seemed to bear no fruit; I wasn't able to prove that the union of a chain of partite edge sets needs to be partite again. Of course I will remove this question if I committed a stupid mistake and there is an easy argument showing that partiteness carries through unions of chains.
