Is the set of all infinite sequences of natural numbers which are eventually constant countable? I know that the set of all infinite sequences with finite length is countable, this set seems like its just countable copies of the set of all infinite sequences with finite length. How to show this rigorously?
 A: Given such a sequence $a:\mathbb{N}\to \mathbb{N}$ define, 
$$ f(a) = (a_0,a_1,a_2,...,a_k,c) $$
(I live is up to you to figure out what that means above). 
Thus, we have constructed an injection $f:\text{const}\to F$, where $F$ are the finite sequences of the natural numbers and $\text{const}$ are the eventually constant sequences. 
A: If you're already at the point where you accept that the finite sequences are countable, then you can simply say that the eventually constant sequences are accounted for by taking any finite sequence and then concatenating it with an infinitely repeated value.
I.e, every eventually constant sequence is equal to some $s^\frown x$ where $s\in\mathbb{N}^{<\mathbb{N}}$ and $x=\langle n,n,n,\ldots\rangle$ for some $n\in\mathbb{N}$.
Thus, you've got an injection from the eventually constant sequences into $\mathbb{N}^{<\mathbb{N}}\times \mathbb{N}$, which is countable since $\mathbb{N}^{<\mathbb{N}}$ and $\mathbb{N}$ are countable.
A: Let 
$$
A_n=\left\{(a)\in \Bbb N^{\Bbb N}\middle|\forall k\in\Bbb N: |a_k|\le n\land a_{n+k}=a_n\right\}
$$
be the set of sequences bounded by $n$ and constant after element index $n$ (earlier included). Then all $A_n$ are finite and $A_n\subset A_{n+1}$ and the requested set is the limit $\bigcup_{n\in \Bbb N}A_n$ of this nested sequence.
