Proof for sequentially continuous function with a first countable domain is continuous Assuming the axiom of countable choice any function $f$ from a first countable space $X$ to $Y$ that is sequentially continuous is necessarily continuous. The gist of the proof I was thinking is that given any $x \in X$ and any neighborhood of $f(x)$ I can then find an open set $S$ contained in that neighborhood. Then I consider the preimage of that set. Assuming by contradiction that the preimage is not a neighborhood then for all $N_i$ of the countable neighborhood basis of $x \in X$ there must be at least one element that is not contained $f^{-1}(S)$ (This is where I'm assuming the axiom of countable choice is used). I form a sequence by induction where for $n$ th element in the countable basis I pick an element $x_n$ in the finite intersection of $N_1,N_2, ..., N_n$-$f^{-1}(S)$ which is non empty since the finite intersection of neighborhood is a neighborhood. This sequence must converge to $x$ but does not converge to $f(x)$ thus proving by contradiction that there must be a neighborhood of $x$ contained in the preimage of $S$. Is this correct? Is there a better approach?
 A: The easiest is to use a characterisation of closed sets that holds in first countable spaces $X$:

$A \subseteq X$ is closed iff for all sequences $(a_n)$ such that all $a_n \in A$ and $a_n \to x$ (in $X$) we have that $x \in A$. The left to right implication holds in all spaces, the right to left only in so-called sequential spaces, of which first countable are an important example.

Now if $C \subseteq Y$ is closed, then to see that $f^{-1}[C]$ is closed, let $(c_n)$ be a sequence all of whose members are in $f^{-1}[C]$ and such that $c_n \to p \in X$. The latter implies by sequential continuity that $f(c_n) \to f(x)$ and $c_n \in f^{-1}[C]$ means $f(c_n) \in C$ for all $n$. 
Now by the always-valid left to right implication, we know from the closedness of $C$, that $f(p) \in C$, and this means that $p \in f^{-1}[C]$. Now we conclude that $f^{-1}[C]$ is closed (as it obeys the condition and is a subset of a sequential space), and we're done showing continuity.
A: To answer aldo decristo's question:
The axiom of countable choice is used in the direction $\Leftarrow$.
Prove by contradiction. Suppose that $A$ is not closed. Then there
exists $x\in\bar{A}\setminus A$. Since the topological space $X$
is first countable, $x$ has a countable base $\{U_{n}\mid u\in\mathbb{N}\}$
(in the sense that $U_{n}$ is an open neighborhood for $x$, and
for each open neighborhood $V$ of $x$, there exists $n$ such that
$U_{n}\subseteq V$). Moreover, by replacing $U_{n}$ with $\cap_{k=1}^{n}U_{k}$
if necessary, without loss of generality, we may assume that $U_{1}\supseteq U_{2}\supseteq\ldots$
Since $x\in\bar{A}$, $U_{n}\cap A\neq\emptyset$. Consider the family
of non-empty sets $\{U_{n}\cap A\mid n\in\mathbb{N}\}$. By the axiom
of countable choice, there exists a function $\theta:\mathbb{N}\rightarrow\cup_{n}(U_{n}\cap A)$
such that $\theta(n)\in U_{n}\cap A$ for each $n$. That $\theta=(\theta(n))$
is a sequence in $A$. Obviously, $\theta(n)\rightarrow x$. For,
let $V$ be an open neighborhood of $x$, then there exists $n_{0}$
such that $U_{n_{0}}\subseteq V$. For any $n\geq n_{0},$we have
$\theta(n)\in U_{n}\subseteq U_{n_{0}}\subseteq V$. This shows that
$\theta(n)\rightarrow x$. By the assumption on the right, we have $x\in A$, which is a contradiction.
