Proof for sequentially continuous function with a first countable domain is continuous

Question

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?

Answer 1

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.