Are locally metrizable topological spaces sequential?
Let $X$ be locally metrisable.
If $A$ is sequentially closed, then if $x \in \overline{A}$, we'd have a metrisable neighbourhood $U_x$ of $x$ by assumption. It would easily follow that there is a sequence $a_n \in A\cap U_x$ so that $a_n \to x$ and sequential closedness of $A$ would imply $x \in A$ and hence $A$ is closed.
The essence of the proof is that a locally metrisable space is first countable.
So indeed sequential continuity with domain $X$ would imply ordinary continuity e.g. (a standard consequence of being a sequential space).