Let $X$ be a topological space and let $Y=[0,1]\subset \mathbb R$ with the ordinary euclidean metric. Now consider the space of continuous functions $\mathcal C(X,Y)$ and endow it with the topology of compact convergence (which is the topology of the compact uniform convergence for sequences).
Is it true that the compact subsets of $\mathcal C(X,Y)$ are also sequentially compact? Without futher assumptions on $X$ I don't think that this is true. Maybe one needs $X$ to be sequentially compact as well.
If $X=\Bbb R$ in the discrete topology, $C(X,Y)$ in the compact convergence topology is just the standard product $\Bbb R^X$. In that space $\{0,1\}^X$ is compact but not sequentially compact, while there are also subspaces like the $\Sigma$-product that are sequentially compact but not compact.
To make it true, we could e.g. assume $X$ is locally compact and hemi-compact so that $C(X,Y)$ becomes a metrisable space for which equivalence of sequential compactness and compactness is classical.
I'm not sure if necessary and sufficient conditions are known for $X$ for this to hold. It would be like an Eberlein-Smulian theorem but for the compact-open topology instead of the weak topology.