Equivalence of continuous and sequential continuous implies first-countable? It is an immediate result that a map from a first-countable space is continuous iff it is sequentially continuous. I was wondering if the converse was also true. That is, is it true that if every map from a space $X$ is continuous iff it is sequentially continuous, then $X$ must be first-countable?
 A: As noted in the comment, the property is equivalent to $X$ being a sequential space. 

Corrected: Suppose that $X$ is not a sequential space. Then there is a sequentially closed $F\subseteq X$ that is not closed. Let $Y=\{0,1\}$ with the topology $\tau=\{\varnothing,\{0\},Y\}$, and let
$$f:X\to Y:x\mapsto\begin{cases}
1,&\text{if }x\in F\\
0,&\text{if }x\in X\setminus F\;.
\end{cases}$$
Then $f$ is sequentially continuous, since $f^{-1}[\{0\}]$ is sequentially open in $X$. However, $f$ is not continuous, since $f^{-1}[\{1\}]=F$ is not closed in $X$.

Now suppose that $X$ is sequential. If $f:X\to Y$ is not continuous, let $U$ be an open set in $Y$ such that $f^{-1}[U]$ is not open in $X$. Since $X$ is sequential, there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $X\setminus f^{-1}[U]$ that converges to some $x\in f^{-1}[U]$, so $f$ is not sequentially continuous. Conversely, if $f:X\to Y$ is not sequentially continuous, there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $X$ converging to some $x\in X$ even though $\langle f(x_n):n\in\Bbb N\rangle$ does not converge to $f(x)$. Thus, there is an open nbhd $U$ of $f(x)$ such that $A=\{n\in\Bbb N:f(x_n)\notin U\}$ is infinite. Let $H=f^{-1}[U]$. Then $x\in H$, and $\langle x_n:n\in A\rangle$ is a sequence in $X\setminus H$ converging to $x$. $X$ is sequential, so $H$ cannot be open, and $f$ is therefore not continuous. Thus, continuity is equivalent to sequential continuity for maps whose domain is a sequential space.
