Two questions on infinite hypergraphs The famous De Bruijn–Erdős theorem and its hypergraphs generalization states the following.
Theorem. Let $V$ be a set, and $E\subset2^V$ be a family of its subsets. Assume that every $e \in E$ is finite and that for some $k \in \mathbb{N}$, every finite sub-hypergraph of $H=(V,E)$ can be properly colored with $k$ colors. Then $H$ can also be properly colored with $k$ colors.
It looks like that the standard proof based on Tychonoff's theorem doesn't work when the edges are allowed to be infinite. So, I wonder if the following analogue of De Bruijn–Erdős theorem holds.

Let $V$ be a set, and $E\subset2^V$ be a family of its subsets. Assume that every $e \in E$ is countable. Moreover, assume that for some $k \in \mathbb{N}$ and for every countable $W \subset V$, the hypergraph $H(W)= (W,E(W))$ can be properly colored with $k$ colors, where $E(W) = \{e \in E: e \subset W\}$. Then $H=(V,E)$ can also be properly colored with $k$ colors.

Erdős and Hajnal proved in [EH] the following related result on families of bounded intersections.
Theorem. Let $V$ be a set, and $E\subset2^V$ be a family of its subsets. Assume that every $e \in E$ is countably infinite. Moreover, assume that there exists $m \in \mathbb{N}$ such that $|e_1\cap e_2|<m$ for all $e_1,e_2 \in E$. Then $H=(V,E)$ can be properly colored with $2$ colors.
Can this result be strengthened as follows to deal with all families of finite intersections?

Let $V$ be a set, and $E\subset2^V$ be a family of its subsets. Assume that every $e \in E$ is countably infinite, and that $e_1\cap e_2$ is finite for all $e_1,e_2 \in E$. Then $H=(V,E)$ can be properly colored with $2$ colors.

[EH] P. Erdős & A. Hajnal, On a property of families of sets, Acta Math. Academiae Scientiarum Hungarica, 12, 87–123 (1964).
 A: To your second question, the answer is no, see problem 18.19 in P. Komjáth, V. Totik:
Problems and Theorems in Classical Set Theory where they cite G. Elekes, G. Hoffmann: On the chromatic number of almost
disjoint families of countable sets, Coll. Math. Soc. J. Bolyai, 10 Infinite and
Finite Sets, Keszthely (Hungary), 1973, 397–402 as a source. (I have this latter book in my office, in case you want it.)
A: The answer to the first question is also no, by a minor modification of the proof of the Elekes-Hoffmann result cited in the answer to the second question. In fact, we get the following:
Theorem There is a set $V$ of cardinality $2^{\aleph_0}$ and a collection $E$ of countably infinite subsets of $V$ such that, for every countable $W \subset V$, $H(W) = (W, E(W))$ is 2-colorable, but $H=(V,E)$ cannot be properly colored with countably many colors.
Proof Let $V$ be the collection of all functions of the form $f:\beta \rightarrow \omega$, where $\beta$ is a countable ordinal. For each such $f$, let $A_f$ be the set of all $n < \omega$ such that the preimage $f^{-1}(\{n\})$ is infinite. For each $n \in A_f$, let $e_{f,n} := \{f\} \cup \{f \restriction \alpha \mid \alpha \in f^{-1}(\{n\})\}$. Then let $E_f := \{e_{f,n} \mid n \in A_f\}$. Note that each element of $E_f$ is countably infinite and has a maximal element, namely $f$ itself. Finally, let $E := \bigcup_{f \in V} E_f$.
Let us first show that $H(W)$ is 2-colorable for every countable $W \subset V$. The point is that, for each such $W$, $E(W)$ is countable, since, by construction, $E(W) \subseteq \bigcup_{f \in W} E_f$, and each $E_f$ is countable. Therefore, one can enumerate both $W$ and $E(W)$ in order-type $\omega$ and use these enumerations to inductively define a coloring $c:W \rightarrow \{0,1\}$, diagonalizing against the elements of $E(W)$ to make sure none of them are monochromatic for $c$ (here we use the fact that every element of $E(W)$ is infinite).
We finally show that $H$ cannot be properly colored with countably many colors. Fix a function $c:V \rightarrow \omega$. We will show that it is not a proper coloring of $H$. Let us define a function $g:\omega_1 \rightarrow \omega$ by recursively specifying $g \restriction \beta$ for $\beta \leq \omega_1$. If $\beta \leq \omega_1$ is a limit ordinal and we have specified $g \restriction \alpha$ for all $\alpha < \beta$, then $g \restriction \beta = \bigcup_{\alpha < \beta} g \restriction \alpha$. For the successor step, suppose that we have constructed $g \restriction \beta$. To define $g \restriction (\beta + 1)$, we must define $g(\beta)$; do so by setting $g(\beta) = c(g \restriction \beta)$.
Let $A$ be the set of all $n < \omega$ such that $g^{-1}(\{n\})$ is unbounded in $\omega_1$, and let $\beta < \omega_1$ be large enough so that:

*

*for all $\alpha < \omega_1$, if $g(\alpha) \notin A$, then $\alpha < \beta$; and

*for all $n \in A$, $g^{-1}(\{n\}) \cap \beta$ is infinite.

Let $n = g(\beta)$, and let $f = g \restriction \beta$. By construction, we have $n = c(g \restriction \beta) = c(f)$. By the first bullet point above, we know that $n \in A$. By the second bullet point, we know that $f^{-1}(\{n\})$ is infinite, so $n \in A_f$ and $e_{f,n} = \{f\} \cup \{f \restriction \alpha \mid \alpha \in f^{-1}(\{n\})$ is in $E$. But for every $\alpha \in f^{-1}(\{n\})$, we have $c(f \restriction \alpha) = c(g \restriction  \alpha) = g(\alpha) = n = c(f)$, so $e_{f,n}$ is monochromatic for $c$, showing that $c$ is not a proper coloring of $H$.
