compactness of thes sequence set

Question

Let $S$ be a compact (in the usual topology) subset of $\mathbb R^n$, let $W = \{(q_k)_{k\in\mathbb{N}}\,\mid\, q_k\in S\}$ be the set of all the sequences taking elements in $S$, let $(f_k)_{k\in\mathbb N}$ be a sequence of continuous functions $f_k:W\to\mathbb R_{\ge 0}$.
For a fixed $\epsilon>0$ consider the set $U_\epsilon=\{q\in W\,\mid\,f_k(q)\ge \epsilon, \forall k\in\mathbb N\}$. I'm wondering what can I conclude about the topological properties of $U_\epsilon$.
My claim is that $U_\epsilon$ is compact, but I'm very weak in topology and my probably terrible attempt to prove that is the following.
$W$ is compact in the product topology, being in a bijection with $\prod_{n\in\mathbb N} S$, hence since $W\subset S$ it suffices to show that $S$ is closed.
for a fixed $k$ one has that $f_k(U_\epsilon) = [\epsilon, A_k]$ for some constant $A_k$. This should come from the continuity of $f_k$ and the compactness of $W$.
But this also means (isn't it?) that $f^{-1}_k([\epsilon,A_k])$ is closed, and if I'm not wrong (which is quite probable) $U_\epsilon = \cap_{k\in\mathbb N} f_k^{-1}([\epsilon,A_k])$, and thus since $U_\epsilon$ is the intersection of (infinitely many) closed sets it is closed and hence compact.
Does this make sense to you?