Consider the continuous function $f:X\to[0,1]$ where $X$ is a compact countable subset of some euclidean space.
We remind the definition of inf-compactness : a function $f$ is inf-compact if for each $\lambda\in\mathbb{R}$ the set $\{x\in X|f(x)\leq \lambda\}$ is compact.
I am wondering if from continuity of $f$ and definition of $X$ we can derive that $f$ is inf-compact ?
My guess is that as $X$ is countable we can endow $X$ with discrete topology but I don't know how to proceed next.
The fact that $X$ is countable or a subset of Euclidean space is not relevant. Because $f$ is continuous, for any $\lambda\in\Bbb R$ we have that: $$\{x\in X:f(x)\le\lambda\}=f^{-1}((-\infty,\lambda])$$Is closed. As a closed subset of a compact space, it is also compact. So $f$ is $\inf$-compact.