Prove that a set of all functions $f:\mathbb N \rightarrow \mathbb N$, which are constant from a certain point, (i.e. a set $\{f\in\mathbb {N^N} \:| \: \exists l\: \exists n \in \mathbb N, \forall k>n, \: f(k)=l\} $) has the same cardinality as $\mathbb{N}$.
I have a big problem with this task, because I thought that if $f\in\mathbb {N^N}$, then this set (let's call it $X$) can have the same cardinality as $\mathbb R$, not $\mathbb{N}$. Because if $|\mathbb R|=|\mathbb {N^N}|$, then how it is possible that $|X|=|\mathbb{N}|$?
Your set $X$ is not $\mathbb{N}^\mathbb{N}$; it's a subset. So, there is no contradiction.
For each $n\in\mathbb N$, let $X_n=\{f\in\mathbb{N}^\mathbb{N}\,|\,m\geqslant n\Longrightarrow f(m)=f(n)\}$. Then $X=\bigcup_{n\in\mathbb N}X_n$ and therefore all you need is to prove is that $|X_n|=|\mathbb{N}|$. That is easy, since $|X_n|$ is equal to the cardinal of the set of all functions from $\{1,2,\ldots,n\}$ to $\mathbb N$, which is equal to the cardinal of $\mathbb N$.