Show that a topological space $X$ is compact if and only if $X^\Bbb N$ is locally compact.
Assume that $X$ is compact, then for every open cover of $X$ there exists a finite subcover. To show that $X^\Bbb N$ I need to show that for any $f \in X^\Bbb N$ there exists a neighborhood $U_f$ such that $\overline{U_f}$ is compact.
I think that $X^\Bbb N$ has the product topology so instead of working with open sets I think I should be working with the basic open sets.
If $U_f$ is any neighborhood of $f \in X^\Bbb N$, then there exists a basic open set $B=\bigcap_{n \in F} \pi_n^{-1}(V_n)$ (where $F \subset \Bbb N$ is finite) such that $$f \in B \subset U_f$$
Can I show that $\overline{B}$ is compact to conclude the result? If so then I would need to show that $\overline{\bigcap_{n \in F} \pi_n^{-1}(V_n)}$ is compact or equivalently that $\overline{\prod_{n \in \Bbb N} V_n}$ where $V_n \ne X$ for only finitely many $n$.
I know that the latter gives me something as $$\overline{\prod_{n \in \Bbb N} V_n} = \prod_{n \in \Bbb N} \overline{V_n}$$ but I cannot figure out how to use the data that $X$ is compact in the proof. I think that I can find an open set in the finite cover of $X$ for every $V_n$?