Let $X$ be a non-empty set such that for each natural $n \in \omega$, does not exists a function with domain $n$ and image $X$. Let $X$ be a non-empty set such that for each natural $n \in \omega$, does not exists a function with domain $n$ and image $X$.
I'm trying to proof that exists $f: \omega \to X$ and injective function.
My attempt:
By the hypothesis, we have that, for each $n \in \omega$, exists $n \to X$ 1-to-1, i.e $n \leq X$ ($X$ dominates $n$).Hence, 
$$ \mathcal F_n = \left \{ f:n \to X \, / \, \text{f is injective}  \right \} \neq \emptyset\ , \quad \forall n \in \omega. $$
Then, $\mathcal F = \bigcup_{n\in \omega} \mathcal F_n \neq \emptyset$. 
Definig the order $f \leq g \iff (dom\, f \subset \, dom\, g)\, \text{and} \,(g|_{dom \, f} = f)$ in $\mathcal F$, we can apply Zorn's Lemma. Then, exists $g: A \subset \omega \to X$ an injection which is maximal in $\mathcal F$.
If $g$ is not surjective, exists $y\in X$ s.t. $g(n) \neq y, \, \forall n \in A$. Suppose that $A \neq \omega$, then $\exists x \in \omega \setminus A$. In this case, $g \cup \{ (x,y) \}$ is an injection function and is greater than $g$, a contradiction. Hence, $A = \omega$ and we are done.
However, I couldn't find a decent argument for the case that $g$ is surjective.
Help?
 A: In order to apply Zorn's lemma, we require that every totally ordered subset of $\mathcal{F}$ has an upper bound. In fact, if the desired conclusion holds, i.e. if there exists an injective $g: \omega\to X$, then the set $\{g\vert_1, g\vert_2, ...\}\subseteq \mathcal{F}$ has no upper bound, for if it had an upper bound $h$, then we would have $n\subseteq \mathrm{dom}(h)$ for each $n$, i.e. $\mathrm{dom}(h) \geq \omega$, and so $h\notin \mathcal{F}$. 
You should define
$$\mathcal{F}_\alpha = \{f: \alpha\to X\,\vert \, f \mathrm{\,\,is\,\, injective}\}$$
for each $\alpha \leq \omega$, not just $\alpha < \omega$. Then let $\mathcal{F} = \bigcup_{\alpha \leq \omega} \mathcal{F}_\alpha$. In addition to running your original argument, you should prove that each totally ordered subset of $\mathcal{F}$ has an upper bound (namely, the union over that set). 
However, the problem you identified still arises. Let $g$ be our maximal element of $\mathcal{F}$. Then $g$ does not have domain $n\in \omega$, for by hypothesis, if this were the case, $g$ would not be surjective, so we'd be able to find a greater element of $\mathcal{F}$, as you showed. Thus, $g$ has domain $\omega$, so we're done. 
