Countability of a sequence of natural numbers

First let's prove that $|\mathbb{N}^2| = |\mathbb{N}|$. An easy way to do this is by using Schröder-Cantor-Bernstein. We easily see that $|\mathbb{N}| \leq |\mathbb{N}^2|$ by sending $n$ to $(n, 0)$. We also find an injection $|\mathbb{N}^2| \leq |\mathbb{N}|$ by sending $(a, b)$ to $2^a 3^b$. By induction we thus have that $|\mathbb{N}^n| = |\mathbb{N}|$ for each natural $n \geq 1$.

Now we can complete the argument based on your sketch.

A finite sequence of length $n$ is just an element of $\mathbb{N}^n$. An eventually constant sequence is a finite sequence with an infinite tail that has constant value $k \in \mathbb{N}$. So for each $n,k \in \mathbb{N}$ there is a set $S_{n,k}$ of sequences that after $n$ values just become the constant sequence with value $k$. Then $$ |S_{n,k}| = |\mathbb{N}^n| = |\mathbb{N}|. $$ We have $$ T = \bigcup_{n,k \in \mathbb{N}} S_{n,k}, $$ which is a countable union of countable sets and hence countable.

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As pointed out in the comments below your question, we generally need the axiom of choice (or a weak version of it) to prove that the countable union of countable sets is countable. This is because for each of the sets in the union we would have to choose a bijection. However, we can get around choice by explicitly constructing bijections $\mathbb{N}^n \to \mathbb{N}$. See for example the Cantor pairing function.

Edit: or see the excellent answer of String, describing an explicit bijection directly, showing that we do not need choice.