Continuity and the Axiom of Choice
Solution 1:
Part of your observation is indeed correct. The axiom of choice is needed, although not in its entire power.
The first model of ZF without the axiom of choice, known as Cohen's first model, is one where there exists an infinite Dedekind-finite set of reals.
We say that a set is Dedekind-finite if it has no infinitely-countable subset, indeed it is consistent without the axiom of choice that such infinite sets exist, and Cohen's model is an example of a model in which such set exists within the real numbers, let us call it $D$.
First we observe that if $a_n$ is a sequence of elements from $D$ then as a set $\{a_n\mid n\in\mathbb N\}$ is finite, otherwise we would have a countably infinite subset of $D$. If so every weakly decreasing/increasing sequence in $D$ is eventually constant. It also holds that there is at least one accumulation point of $D$ (within $D$, that is). Otherwise we could have separated all the points by disjoint intervals and would then reach the conclusion that $D$ is at most countable.
Let $S=D\setminus\{a\}$ for some accumulation point $a$ in $D$. Now consider $f\colon D\to\mathbb R$ define as: $$f(x)=\begin{cases}1 & x=a\\ 0 &x\neq a\end{cases}$$ This function is not continuous at $a$ for obvious reasons, for $\varepsilon=\frac12$ we have that for $\delta>0$ we have some $d\in S$ such that $|d-a|<\delta$ we still have $|f(d)-f(a)|=1>\varepsilon$.
However, given a sequence $a_n$ whose limit is $a$ then the sequence is eventually constantly $a$, therefore $f(a_n)=f(a)=1$ for almost all $n$.
Note that if we assume that $f\colon\mathbb R\to\mathbb R$ is continuous at $x$, and $x_n$ is a sequence approaching $x$ then indeed $f(x_n)$ approaches $f(x)$.
Suppose not, then there is some $\varepsilon>0$ that for large enough $n$ we have $|f(x)-f(x_n)|>\varepsilon$, then for every $\delta>0$ we can find $n$ such that $|x-x_n|<\delta$, which contradicts the $\varepsilon$-$\delta$ continuity at $x$.
On the other hand, assuming that for a given point $x$ continuity at $x$ equals sequential continuity there is far from the axiom of choice. Why? Note that we really just need countable choice to assert the equivalence, and it is quite simple to construct models in which the axiom of choice does not hold but the axiom of countable choice holds.
Indeed we know that the assertion:
$f\colon\mathbb R\to\mathbb R$ is continuous at a point $x$ if and only if it is sequentially continuous at that point.
Is equivalent to the axiom of countable choice for subsets of the real line; and I suppose we can use that to deduce the same for complex valued functions quite easily.
I should also add the following:
If the equivalence between the two forms of continuity does not hold it means that the axiom of countable choice does not hold. In turn we cannot ensure that countable unions of countable sets are still countable; this allows great difficulties in ensuring that real analysis works as we are familiar with it.
However, there are interesting models for analysis where the axiom of choice does not hold. Solovay's model in which every set is Lebesgue measurable (and much much more) is such model. Just today I ran into a relatively old paper speaking excitingly about "Solovayan functional analysis" which will be a new and exciting field.
With gratitude to Brian pointing out that continuity everywhere is still equivalent to sequential continuity everywhere; but not when we require a function to be continuous at a certain point (the counterexample in the Dedekind-finite set would be a function which is not continuous everywhere).
Solution 2:
The implication $(1)\to (2)$ can be proved in ZF alone, though this requires major revision of the argument.
Assume that $f:\Bbb R\to\Bbb R$ is sequentially continuous (i.e., satisfies (1)). Let $x\in\Bbb R$ be arbitrary.
Claim: $f\upharpoonright(\Bbb Q\cup\{x\})$ is continuous at $x$.
Proof: Enumerate $\Bbb Q=\{q_n:n\in\omega\}$. If $f\upharpoonright(\Bbb Q\cup\{x\})$ is not continuous at $x$, there is an $\epsilon>0$ such that for each $k\in\omega$, $$A_k\triangleq\{q\in\Bbb Q:|q-x|<2^{-k}\text{ and }|f(q)-f(x)|\ge\epsilon\}\ne\varnothing\;.$$ Let $$n(0)=\min\{k\in\omega:q_k\in A_0\}\;.$$ Given $n(m)$, let $$n(m+1)=\min\{k\in\omega:k>n(m)\text{ and }q_k\in A_{m+1}\}\;.$$ Then $\langle q_{n(k)}:k\in\omega\rangle\to x$, but $|f(q_{n(k)})-f(x)|\ge\epsilon$ for all $k\in\omega$, which is a contradiction. Note that no choice was used in this construction. $\dashv$
Now let $\epsilon>0$. It follows from the Claim that there is a $\delta>0$ such that $|f(x)-f(q)|\le\epsilon$ whenever $q\in\Bbb Q$ and $|x-q|\le\delta$. Now suppose that $y\in\Bbb R$ with $|x-y|\le\delta$. Let $I$ be the closed interval whose endpoints are $x$ and $y$. For each $k\in\omega$ let $$n(k)=\min\{m\in\omega:q_m\in I\text{ and }|q_m-y|<2^{-k}\}\;.$$ Then $\langle q_{n(k)}:k\in\omega\rangle\to y$, so $\langle f(q_{n(k)}):k\in\omega\rangle\to f(y)$. However, each $q_{n(k)}\in I$, so $|q_{n(k)}-x|\le\delta$ for each $k\in\omega$, and therefore $|f(q_{n(k)})-f(x)|\le\epsilon$ for each $k\in\omega$. Clearly this implies that $|f(y)-f(x)|\le\epsilon$ as well.
Thus, we’ve shown that for each $\epsilon>0$ there is a $\delta>0$ such that $|f(y)-f(x)|\le\epsilon$ whenever $|y-x|<\delta$, which is sufficient.
This argument is expanded from the ZF proof of Theorem 3.15 in Horst Herrlich, Axiom of Choice, Lecture Notes in Mathematics 1876. Note that it is not true in ZF that a function $f:\Bbb R\to\Bbb R$ is continuous at a point $x$ iff it is sequentially continuous at $x$: in Theorem 4.54 he proves that this assertion is equivalent to $\mathbf{CC}(\Bbb R)$, the assertion that every countable family of non-empty subsets of $\Bbb R$ has a choice function.
Added: I failed to notice that the functions in the original question are complex-valued, not real-valued. However, the argument can easily be adapted to $\Bbb R^2$, replacing $\Bbb Q$ by $\Bbb Q^2$ and using the $\max$ norm.
Added2: The only slightly tricky bit is figuring out what to use for $I$. Suppose that $x=\langle x_1,x_2\rangle,y=\langle y_1,y_2\rangle\in\Bbb R^2$ with $0<\|x-y\|\le\delta$. If $x_1\ne y_1$ and $x_2\ne y_2$, let $$I=\big[\min\{x_1,y_1\},\max\{x_1,y_1\}\big]\times\big[\min\{x_2,y_2\},\max\{x_2,y_2\}\big]\;.$$ If $x_1\ne y_1$ and $x_2=y_2$, let $$I=\big[\min\{x_1,y_1\},\max\{x_1,y_1\}\big]\times\big[x_2,x_2+\delta\big]\;,$$ and if $x_1=y_1$ and $x_2\ne y_2$ let $$I=\big[x_1,x_1+\delta\big]\times\big[\min\{x_2,y_2\},\max\{x_2,y_2\}\big]\;.$$
Let $\Bbb Q^2=\{q_n:n\in\omega\}$ be an enumeration of $\Bbb Q^2$, and for $k\in\omega$ let $$n(k)=\min\{m\in\omega:q_m\in I\text{ and }\|q_m-y\|<2^{-k}\}\;.$$ Everything else is the same as before, except that $|\cdot|$ must be replaced throughout by $\|\cdot\|$, where $$\|\langle x,y\rangle\|=\max\{|x|,|y|\}\;.$$
With very minor modification this works for $\Bbb R^n$ for any $n\in\Bbb Z^+$.