¿A Dedekind-infinite set $X$ has a countable subset, and a proper subset equipotent to $X$, but both with intersection empty? In a more detailed way,
Definition: A set $X$ is said to be Dedekind-infinite if it has a proper subset $A$ equipotent (similar/in bijection) to $X$.
Let $X$ be a Dedekind-infinite set. Does it exists  $A, B$, subsets of $X$ such that,  $A\sim X$, $B\sim \mathbb{N}$ and $A\cap B=\emptyset$?
If not, what conditions are needed to ensure this?
In this case $U\sim V$ means "There is a bijective function $f:U\rightarrow V$".
The question comes to me and I think it's true; I can't find a counterexample.
Also, I think, the proposition is intuitive. The set of natural numbers is the "least" inductive set, and then, the first infinite set, and it has that property.

Thanks in advance
 A: The answer is yes. The key observation is that any countably infinite set $A$ can be partitioned into two countably infinite subsets $A=A_1\sqcup A_2$ - this is just Hilbert's Hotel - and of course we then have $A\sim A_1\sim A_2$.
So suppose $X$ is Dedekind-infinite. I claim that $X$ has a countably infinite subset $A$. Before proving this, let's see why this is helpful: if $A\subset X$ is countably infinite, write $A=A_1\sqcup A_2$ as above. Then $A_2\cup (X\setminus A)\sim X$ (why? because $A_2\sim A$!), $A_1\sim\mathbb{N}$, and $[A_2\cup (X\setminus A)]\cap A_1=\emptyset$.
So now we just have to show that a Dedekind-infinite set has a countably infinite subset. Let $f: X\rightarrow X$ be injective but not surjective, and let $x\in X\setminus ran(f)$. Now think about the sequence $x, f(x), f(f(x)), f(f(f(x)))$, ... It's a good exercise to show that this sequence never repeats - that is, for $m\not=n$ we have $f^m(x)\not=f^n(x)$. So this defines a countably infinite subset of $X$, namely $\{f^n(x): n\in\mathbb{N}\}$.
