In fact, every rainbow hypergraph has chromatic number $2$.

Let $H=(V,E)$ be a rainbow hypergraph, $E=\{e_2,e_3,\dots\}$, $|e_n|=n$. Consider a random coloring $c:V\to\{0,1\}$, let $A$ be the event that $c$ is not a proper coloring of $H$, and let $A_n$ be the event that $c$ is constant on $e_n$. Then $$P(A)=P\left(\bigcup_{n=2}^\infty A_n\right)\lt\sum_{n=2}^\infty P(A_n)=\sum_{n=2}^\infty\frac1{2^{n-1}}=1,$$ so proper $2$-colorings exist and $\chi(H)=2$.

P.S. Here is an alternative argument which even proves a slightly stronger result: a rainbow graph remains $2$-colorable if one more edge is added arbitrarily.; i.e., there are now two edges of size $2$ and one edge of size $n$ for each integer $n\gt2$.

Let $H=(V,E)$ be a hypergraph, $E=\{e_1,e_2,e_3,\dots\}$ where $|e_n|=\max(n,2)$. We color the vertices sequentially, coloring $2$ vertices at each step, one red and the other blue; this is done in such a way that after the $n^\text{th}$ step is completed (if not sooner) the edge $e_n$ contains at least one vertex of each color. This can always be done, unless the elements of $e_n$ have all been given the same color before the $n^\text{th}$ step; but that can't happen because $|e_n|\ge n$ and only $n-1$ vertices have received each color.

In fact, every rainbow hypergraph has chromatic number $2$.

Let $H=(V,E)$ be a rainbow hypergraph, $E=\{e_2,e_3,\dots\}$, $|e_n|=n$. Consider a random coloring $c:V\to\{0,1\}$, let $A$ be the event that $c$ is not a proper coloring of $H$, and let $A_n$ be the event that $c$ is constant on $e_n$. Then $$P(A)=P\left(\bigcup_{n=2}^\infty A_n\right)\lt\sum_{n=2}^\infty P(A_n)=\sum_{n=2}^\infty\frac1{2^{n-1}}=1,$$ so proper $2$-colorings exist and $\chi(H)=2$.

P.S. Here is an alternative argument which even proves a slightly stronger result: a rainbow graph remains $2$-colorable if one more edge is added arbitrarily.; i.e., there are now two edges of size $2$ and one edge of size $n$ for each integer $n\gt2$.

Let $H=(V,E)$ be a hypergraph, $E=\{e_1,e_2,e_3,\dots\}$ where $|e_n|=\max(n,2)$. We color the vertices sequentially, coloring $2$ vertices at each step, one red and the other blue; this is done in such a way that after the $n^\text{th}$ step is completed (if not sooner) the edge $e_n$ contains at least one vertex of each color. This can always be done, unless the elements of $e_n$ have all been given the same color before the $n^\text{th}$ step; but that can't happen because $|e_n|\ge n$ and each color has only been used $n-1$ times before the $n^\text{th}$ step.

In fact, every rainbow hypergraph has chromatic number $2$.

Let $H=(V,E)$ be a rainbow hypergraph, $E=\{e_2,e_3,\dots\}$, $|e_n|=n$. Consider a random coloring $c:V\to\{0,1\}$, let $A$ be the event that $c$ is not a proper coloring of $H$, and let $A_n$ be the event that $c$ is constant on $e_n$. Then $$P(A)=P\left(\bigcup_{n=2}^\infty A_n\right)\lt\sum_{n=2}^\infty P(A_n)=\sum_{n=2}^\infty\frac1{2^{n-1}}=1,$$ so proper $2$-colorings exist and $\chi(H)=2$.

In fact, every rainbow hypergraph has chromatic number $2$.

Let $H=(V,E)$ be a rainbow hypergraph, $E=\{e_2,e_3,\dots\}$, $|e_n|=n$. Consider a random coloring $c:V\to\{0,1\}$, let $A$ be the event that $c$ is not a proper coloring of $H$, and let $A_n$ be the event that $c$ is constant on $e_n$. Then $$P(A)=P\left(\bigcup_{n=2}^\infty A_n\right)\lt\sum_{n=2}^\infty P(A_n)=\sum_{n=2}^\infty\frac1{2^{n-1}}=1,$$ so proper $2$-colorings exist and $\chi(H)=2$.

P.S. Here is an alternative argument which even proves a slightly stronger result: a rainbow graph remains $2$-colorable if one more edge is added arbitrarily.; i.e., there are now two edges of size $2$ and one edge of size $n$ for each integer $n\gt2$.

Let $H=(V,E)$ be a hypergraph, $E=\{e_1,e_2,e_3,\dots\}$ where $|e_n|=\max(n,2)$. We color the vertices sequentially, coloring $2$ vertices at each step, one red and the other blue; this is done in such a way that after the $n^\text{th}$ step is completed (if not sooner) the edge $e_n$ contains at least one vertex of each color. This can always be done, unless the elements of $e_n$ have all been given the same color before the $n^\text{th}$ step; but that can't happen because $|e_n|\ge n$ and only $n-1$ vertices have received each color.