Using the axiom of choice to prove that "sequential continuity" implies continuity

Question

Suppose that $f$ is defined on an open neighbourhood of a point $a\in\mathbb R$. I wish to prove that if $f$ is "sequentially continuous" at $a$, then $f$ is continuous at $a$. The definition of "sequentially continuous" is that for every sequence $(a_n)$ that approaches $a$ we have $f(a_n)\to f(a)$. The proof is by contraposition.

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Suppose that $f$ is not continuous at $a$. Then, there is an $\varepsilon>0$ such that for every $\delta>0$ there is an $x$ for which $|x-a|<\delta$ but $|f(x)-f(a)|\ge\varepsilon$. We will now construct a sequence $(a_n)$ such that $(a_n)\to a$ but it is not true that $f(a_n)\to f(a)$.

We know that there is an $\varepsilon>0$ and an $x$ such that $|x-a|<1$ and $|f(x)-f(a)|\ge\varepsilon$. Let $a_1$ be one such $x$. We know that there is an $\varepsilon>0$ and an $x$ such that $|x-a|<0.1$ and $|f(x)-f(a)|\ge\varepsilon$. Let $a_2$ be one such $x$. Continue in this fashion. More formally, let $$ A_i=\{x:|x-a|<10^{-i}\text{ and }|f(x)-f(a)|\ge\varepsilon\} $$ and consider the set $$ S=\{A_i:i\in\mathbb N\} $$ The discontinuity of $f$ at $a$ means that each $A_i$ is non-empty, and so we can use the axiom of choice to define a function $g:S\to\bigcup S$ such that $g(A_i)\in A_i$ for all $A_i\in S$. Now let $a_i=g(A_i)$, and we are done.

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My questions are:

1. Is this argument correct?
2. Since $S$ is countable, am I right in thinking that we only need the axiom of countable choice to prove this theorem?