Let $Y$ be a compact subset of $\mathbb{R}$, and $X\subseteq Y^{[0, 1]}$ be the set of non-decreasing functions from $[0,1]$ to $Y$, where $Y^{[0,1]}$ has product topology. Is $X$ compact and first-countable?
I know this is the case when $Y=[0,1]$, as then $X$ is a Helly space. But I don't know if this generalizes. If it does not, what additional assumptions on $Y$ would be required to make this true?
$Y$ is compact, so it is contained in $[-R, R]$ for some $R > 0$. It is then clear that $X$ is a closed subset of the space of non-decreasing functions from $[0, 1]$ to $[-R, R]$. The latter space is clearly homeomorphic to the Helly space, so compact and first-countable. As $X$ is a closed subspace, it must be compact and first-countable as well.