Let $\mathbb{N}$ denote the set of all natural numbers and let $2^{\mathbb{N}}$ denote the collection of all subsets of $\mathbb{N}$. To prove that the collection $2^{\mathbb{N}}$ is uncountable, for a contradiction, we assume that it is countably infinite, so that the subsets of $\mathbb{N}$ can be listed $A_1, A_2, ........$; hence every subset of $\mathbb{N}$ is $A_k$ for some $k\in \mathbb{N}$. Now, the key point in the proof is to define a set $A=\{k\in\mathbb{N}: k\not\in A_k\}$. With this definition we easily conclude the proof.
Here is my question: Since $\emptyset \subset \mathbb{N}$, we see that the set $A$ is non-empty; but, instead of using this fact, to see that $A\neq \emptyset$, is it possible to prove the existence of (hence, to have) an integer $k$ with $k \not\in A_k$ (so that $A\neq \emptyset$) ? Thanks in advance for any comment/answer.
Given an injective map $\Bbb N\to 2^{\Bbb N}$, $k\mapsto A_k$, you cannot assume for any $k$ that $k\notin A_k$. For example, we might have $A_k=\{k\}$ for all $k$. Of course, if you assume that the map is surjective, you can show whatever you want - simply because there is no such surjection and ex falso quodlibet. In fact, for every $B\subseteq \Bbb N$ (not only for $B=\emptyset$), we immediately see that $A\ne B$, because we assume that there exists some $k$ with $A_k=B$ and the diagonal argument shows that $A\ne A_k$.