Why do we need the axiom of choice in this particular case?

Question

I am sorry if this is a stupid question, but I really want to hear an answer from someone more knowledgeable. So the sequential criterion for the limit says that: if $f: A \subseteq \mathbb{R} \to \mathbb{R}$ and $c \in \mathbb{R}$ is a cluster point of $A$ then the following are equivalent:

• $\lim_{x \to c} f(x) = L$
• For every sequence $(x_n)$ in $A \setminus \{c\}$ that converges to $c$ the sequence $(f(x_n))$ converges to L.

The usual proof of this, first proves that (1) implies (2) and then we prove that the negation of (1) implies the negation of (2). My question is about the latter part. First we assume that $\lim_{x \to c} f(x) = L$ is false which means: $$ (\exists \epsilon > 0)(\forall \delta > 0)(\exists x \in A)(0 < | x - c | < \delta \land | f(x) - L | \geq \epsilon). $$ Using existential instantiation, we can call $\epsilon_0$ to one such $\epsilon$. We then build the following sequence $(x_n)$, for each $n \in \mathbb{Z}^+$ we define $x_n$ as a real number for which it is true that $$ (0 < | x_n - c |< \frac{1}{n} \land |f(x_n) - L|\geq \epsilon_0); $$ we know such $x_n$ exists because $\frac{1}{n}$ is positive and so by existential instantiation again, we guarantee one such element. After that we prove $(x_n)$ converges to $c$ but $(f(x_n))$ does not converge to $L$, which concludes the proof.

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My problem is, according to the set theory book I was reading, when we picked for each $n$ a real number $x_n$ using existential instantiation we are actually using the axiom of choice. Why is this the case? I mean we're just using a logic law that should be independent of the axioms of set theory, so why are we assuming the axiom of choice when using existential instantiation an infinite number of times?