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Let ${\{A_n\}}_{n=1}^{\infty }$ be a countable collection of countable sets. Denote $A={\bigcup }_{n=1}^{\infty }{A}_n$. For each set $A_n$, define an injection $f_n:A_n\to \mathbb{N}$. Now define a function $f:A\to \mathbb{N}$ as follows: for each element $x\in A$, let $i$ be the first index such that $x\in A_i$. Define $$f(x)=2^i\cdot 3^{f_i(x)}$$ This ensures that $f$ is an injection from $A$ to $\mathbb{N}$, as the use of distinct powers of 2 for each set guarantees that no two elements from different $A_n$'s are mapped to the same value in $\mathbb{N}$, this follows from the uniqueness of the prime factorization. If $A$ is finite, then it is trivially countable because finite sets are a special case of countable sets. If $A$ is infinite, then the image of $f$, $\text{im}(f)$, is an infinite subset of $\mathbb{N}$, and any infinite subset of $\mathbb{N}$ is countable (since it has a bijection to $\mathbb{N}$). Therefore, $A$ is bijective to $\text{im}(f)$, and thus $A$ is countable.