Is this set uncountable? $A = \{A_n \colon n \in \mathbb{N}\}$ where $A_n$ is the set $\mathbb{N}$ with the number $n$ removed from it

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The set in the title is presented in this answer as an example of a similar set to the $P(\mathbb{N})$ (in the context of explaining the necessity of the axiom of choice in the existence of a well ordering on reals), but the definition of it in terms of enumerating the $\mathbb{N}$ gives me the impression it's actually a countable set.

(This should actually have been a comment to the original answer but I don't have enough reputation. Feel free to remove if I'm violating any guidelines).

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By definition, a set is countable if there exists a bijection from itself to the naturals. In other words, if there is a one-to-one function from $A$ to $\mathbb{N}$ that does not lefts out any natural number. In this case, it is easy to find such a function: $$ f: \mathbb{N} \rightarrow A, \ f(n) = A_n $$ It is easy to show that $f$ is one-to-one and for every $a \in A$, there exists $n_a \in \mathbb{N}$ such that $f(n_a) = a$

In conclusion, the set $A_n$ is countable