Let $X$ be a topological space, $(Y, d)$ a metric space and $C(X, Y)$ the space of continuous maps with the topology of compact convergence.

**Question:** What is a minimal topological condition on $X$ such that $C(X, Y)$ is a sequential space?

The motivation of the question is the following:

Let $(f_n)$ be a sequence in $C(X, Y)$ that is equicontinuous and $ \overline{ \{ f_n(x) : n \in \mathbb{N} \} } $ is compact in $(Y, d)$ for every $x \in X$. By Ascoli theorem, the closure $ \overline{ \{ f_n : n \in \mathbb{N} \} } $ in $C(X, Y)$ is compact. Unfortunately, in general $(f_n)$ may not have any convergent subsequence (in the topology of compact convergence).

To guarantee a convergent subsequence, it is enough for $C(X, Y)$ to be a sequential space. Then $ \overline{ \{ f_n : n \in \mathbb{N} \} } $ would be a compact subspace that is also sequential, which implies that it is sequentially compact.

It is well known that if $X$ is hemicompact, then $C(X, Y)$ is metrizable. Of course $C(X, Y)$ would be sequential in this case. But the hemicompactness of $X$ is too strong a condition. We only need $C(X, Y)$ to be sequential, and metrizability is a big overkill.

On the other hand, it is not enough for $X$ to be a locally compact metric space. See this Math.SE page for a counter-example.