Is a finite union of countable sets at most countable? I thought of this result:

A finite union of countable sets is at most countable.

which I also tried to prove:

Let $A_1, A_2, \dots, A_n$ be a finite collection of countable sets. Then, each $A_i$ must be enumerable, that is, we can write
\begin{align*}
     A_1 &= \tau_1, \tau_2, \dots \\
     A_2 &= \zeta_1, \zeta_2, \dots \\
     &\vdots \\
     A_n &= \phi_1, \phi_2, \dots 
 \end{align*}
We can then construct a sequence as follows:
\begin{equation}\tag{1}
     \tau_1, \tau_2, \dots, \dots, \zeta_1, \zeta_2, \dots, \dots, \dots, \phi_1, \phi_2, \dots
 \end{equation}
Note that some of the terms in the sequence in (1) might be repeated. If an infinite number of terms of this sequence are repeated, then we can retain a single copy of each of the repeated terms and eliminate all the duplicates; this will leave us with a finite number of terms in the sequence in (1). Similarly, if only a finite number of terms in the sequence in (1) are repeated, then eliminating all the duplicates as explained above; we will be left with a sequence of terms from (1) which can still be indexed by set of positive integers. Therefore, $\bigcup_{i=1}^{n}A_i$ is at most countable.

I have two questions. Is it true that a finite union of a countable sets is at most countable? Secondly, are there any inaccuracies in my proof?
 A: The claim is true, and as the comments mention, an even stronger statement is true: a countable union of countable sets is countable (if you want you can speak of "at most countable" everywhere... depending on how exactly you defined "countable").
However, your proposed proof is incorrect, because it is not clear what your sequence actually is. How are you supposed to go "infinitely far" using
\begin{align}
\tau_1, \tau_2, \tau_3, \dots
\end{align}
and then once you "list out infinitely many, start again" with
\begin{align}
\dots \zeta_1, \zeta_2, \zeta_3, \dots?
\end{align}
What are the $\dots$ even supposed to mean? Note that you should only use the three little dots when you're $100\%$ confident that you can translate that intuitive notation into something more rigorous and unambiguous (because the misunderstanding of "$\dots$" is the cause of so many confusions in math).
This is incorrect. One approach to construct a sequence is as follows:
\begin{align}
\tau_1, \zeta_1, \dots, \phi_1| \tau_2, \zeta_2, \dots \phi_2|\tau_3, \zeta_3, \dots, \phi_3, \dots
\end{align}
(I put vertical bars $|$ only to help "see" what I mean; just think of them as a comma if you wish).

If you want to be slightly more precise, you can start by indexing the elements of the set slightly differently: for each $i \in \{1, \dots, n\}$, let the elements of the set $A_i$ be denoted as $A_i = \{a_{ki}\}_{k \in \Bbb{N}}$. Then, the sequence we're defining is as follows
\begin{align}
a_{11}, \dots a_{1n}| a_{21}, \dots, a_{2n}| a_{31}, \dots a_{3n}| \dots
\end{align}
In words, you're "going down vertically $n$ times, then moving to the right to the next column, then going down again, and then repeating".
If you want the countable union case, look up the diagonal argument.
