Consider the claim "for any infinite set $X$, there exists an unbounded $f:X \to \mathbb{R}$". If we assume the axiom of choice, this claim is trivial to prove. Indeed, given choice we know there exists an $S \subset X$ which is countably infinite, and thus we map that set to $\mathbb{N} \subset \mathbb{R}$ and map each $x \in X \setminus S$ to $0$.
Can this be proven without choice?
This is absolutely not provable. If $X$ is an amorphous set, which is an infinite set which cannot be split into two disjoint infinite subsets, then every function from $X$ into a linearly ordered set, in particular $\Bbb R$, has a finite range.
And of course, the existence of amorphous sets is consistent with $\sf ZF$.
More generally, $X$ cannot be mapped onto $\Bbb N$ if and only if $X$ cannot be mapped onto an unbounded subset of $\Bbb R$. And we know that $X$ cannot be mapped onto $\Bbb N$ if and only if its power set is Dedekind-finite.