Product of perfect maps is perfect

Suppose $X=\prod_{s\in S}X_s$ and $Y=\prod_{s\in S}Y_s$ have the product topology and $f_s:X_s\to Y_s$ are functions. If each $f_s$ is perfect then $f=\prod_{s\in S}f_s$ is perfect.

I already proved $f$ is continuous and the fibers $f^{-1}(y)$ are compact. But I haven't proved $f$ is closed yet.

In this answer Finite Product of Closed Maps Need Not Be Closed, it is shown the product of closed maps need not be closed, so the compactness of the fibers must be important, but how? Say $A\subseteq X$ is closed. Which way do I need to follow to prove $f(A)$ is closed? I don't know how to start.

1 Answer

Some quotations from “General Topology” by Ryszard Engelking.