I've skipped over some of the references in my proof for brevity. The following is an exercise from Terence Tao's Analysis I, specifically section 8.1. on countablity. ("countable" here refers to "countably infinte").
Exercise 8.1.9. Suppose that $I$ is an at most countable set, and for each $\alpha \in I$, let $A_\alpha$ be an at most countable set. Show that the set $\bigcup_{\alpha\in I} A_\alpha$ is also at most countable. In particular, countable unions of countable sets are countable. (This exercise requires the axiom of choice, see Section 8.4).
Proof. Suppose $I$ is an at most countable set, and for each $\alpha \in I$ let $A_\alpha$ be an at most countable set.
We can construct injective functions $f\colon I \to \mathbb{N}$ and $g_\alpha\colon A_\alpha \to \mathbb{N}$, since we can either construct a bijection to $\mathbb{N}$ (countable, Defn 8.1.1) or to a subset of $\mathbb{N}$ specifically $\{i\in \mathbb{N}\mid 1\leq i \leq n\}$ for some finite cardinality $n$ (finite, Defin 3.6.5).
For all $x\in \left(\bigcup_{\alpha\in I} A_\alpha \right)$, we know we have some $\alpha \in I$ s.t. $x\in A_\alpha$ (Axiom of Union) hence the set $\{\alpha \in I \mid x \in A_\alpha\}$ is non-empty. Thus, $\min f(\{\alpha \in I \mid x \in A_\alpha\}$ exists and is unique (Well orderng principle), and is within the range of $f$. Hence we have a function $\beta\colon \bigcup_{\alpha \in I} A_\alpha \to I$ s.t. $\beta(x) = f^{-1} \big(\min f(\{\alpha \in I \mid x \in A_\alpha\}\big)$ where $x\in A_{\beta(x)}$.
Let $h\colon \bigcup_{\alpha \in I} A_\alpha\to \mathbb{N}\times \mathbb{N}$ be defined by $h(x) = \Big(f\big(\beta(x)\big), g_{\beta(x)}(x)\Big)$. $h$ can be shown to be injective since $f$ and $g_{\beta(x)}$ are injective (skipped over for brevity). $h\left(\bigcup_{\alpha \in I} A_\alpha\right)$ is at most countable since it is the subset of $\mathbb{N}\times \mathbb{N}$ (Corollary 8.1.7), where $\mathbb{N}\times \mathbb{N}$ is countable (Corollary 8.1.13). Hence, since $h$ is injective, $\bigcup_{\alpha \in I} A_\alpha$ is also at most countable. $\square$
Tao hasn't gone over the Axiom of Choice yet in the book, so I'm not exactly sure what it specifically entails. However, if such axiom is required, it should be apparent from what I know that I am making some leap of logic. I'm not sure what that is, I'm suspicous it has something to do with the construction of $\beta$ which is used to find some unique $\alpha\in I$ for any $x$ s.t. $x\in A_\alpha$. Or perhaps even though $g_\alpha$ exists, it probably requires Axiom of Choice to choose such $g_\alpha$ for potentially infinite $\alpha$. Note: Here I use Well ordering principle specifically for natural numbers which I believe doesn't need Axiom of Choice.
I'm also not 100% sure my proof is correct, so some review there would be appreciated.