In my introductory Analysis course, we learned two definitions of continuity.

$(1)$ A function $f:E \to \mathbb{C}$ is *continuous at* $a$ if every sequence $(z_n) \in E$ such that $z_n \to a$ satisfies $f(z_n) \to f(a)$.

$(2)$ A function $f:E \to \mathbb{C}$ is *continuous at* $a$ if $\forall \varepsilon>0, \exists \delta >0:\forall z \in E, |z-a|<\delta \implies |f(z)-f(a)|<\varepsilon$.

The implication $(2)\implies(1)$ is trivial (though I will happily post a proof if there is sufficient interest). The proof of the implication $(1)\implies(2)$ is worth remarking on, though.

*Proof that* $(1)\implies(2)$*:*

Suppose on the contrary that $\exists \varepsilon>0:\forall \delta>0, \exists z \in E:\left (|z-a|<\delta \; \mathrm{and} \; |f(z)-f(a)|\ge \varepsilon\right )$. Let $A_n$ be the set $\{z\in E:|z-a|<\frac{1}{n} \; \mathrm{ and }\; |f(z)-f(a)|\ge\varepsilon\}$. Now use the Axiom of Choice to construct a sequence $(z_n)$ with $z_n \in A_n \; \forall n \in \mathbb{N}$. But now $a-\frac{1}{n}<z_n<a+\frac{1}{n}\; \forall n \in \mathbb{N}$ so $z_n \to a$. So $f(z_n) \to f(a)$. But $|f(z_n)-f(a)|\ge\varepsilon\; \forall n \in \mathbb{N}$, which is a contradiction.

You will have noticed that the above proof uses the Axiom of Choice (the lecturer didn't explicitly spell out the dependence, but it's definitely there). My question is: is it possible to prove that $(1) \implies (2)$ *without* using the Axiom of Choice. I strongly suspect that it isn't. In that case, can anyone prove that we have to use the Axiom of Choice? I can think of three ways to do this:

(A) Show that $\left( (1) \implies (2)\right)\implies \mathrm{AC}$. ~~I suspect that this statement is untrue.~~ This is definitely untrue, as Arthur points out, because I only used the axiom of countable choice, which is strictly weaker than AC.

(B) Show that $(1)\implies (2)$ is equivalent to some other statement known to require the Axiom of Choice (the obvious example being the well-ordering of the real numbers).

(C) Construct or show the existence of a model of ZF in which there exist sequences which satisfy $(1)$ but not $(2)$.

Of course, if anyone can think of another way, I would be very interested to hear about it.

One final note - I am aware that very many theorems in Analysis use the Axiom of Choice in one way or another, and that this is just one example of such a theorem. If there exists a model of ZF like the one described in (C), is the study of Analysis in that model interesting?