Q:

Closed set - multicolored set of reals

My question refers to the compactness argument on page 5 of this document , and is very simple. How we we know that the sets $C_x$ are closed? I don't understand the explanation they give (a basis for the open sets is the set of all colorings whose values are specified on a finite
number of points). 
I am familiar with the definition of compact space (every cover has a finite subcover), closed sets (their complements are open) and basis (collection of open sets such that any open set in the space can be written as a union of elements in the basis).

A:

It follows from the definition of the product topology on $[k]^{\Bbb R}$.
Let $\mathscr{F}$ be the family of non-empty finite subsets of $\Bbb R$. For each $F\in\mathscr{F}$ and each $k$-coloring $\xi:F\to[k]$ let
$$B(F,\xi)=\left\{c\in[k]^{\Bbb R}:c\upharpoonright F=\xi\right\}\;,\tag{1}$$
the set of all $k$-colorings of $\Bbb R$ that agree with $\xi$ on the finite set $F$. When $[k]$ is given the discrete topology, the product space $[k]^{\Bbb R}$ has as a base the family of all sets of the form $(1)$; this follows immediately from any of the common equivalent definitions of the product topology.
Now let $C_x$ be the set of all $k$-colorings of $\Bbb R$ such that $x+S$ is multicolored. Let $x+S\in\mathscr{F}$. A $k$-coloring $c\in[k]^{\Bbb R}$ belongs to $C_x$ if and only if $c[F]=[k]$. Let $\Xi$ be the set of $k$-colorings $\xi$ of $F$ such that $\xi[F]\ne[k]$; then 
$$[k]^{\Bbb R}\setminus C_x=\left\{c\in[k]^{\Bbb R}:c\upharpoonright F\in\Xi\right\}=\bigcup_{\xi\in\Xi}B(F,\xi)\;.$$
$[k]^{\Bbb R}\setminus C_x$ is therefore open, being a union of open sets, and its complement $C_x$ must be closed. 
(Note, by the way, that $C_x$ is also open, being the union of the basic open sets $B(F,\xi)$ such that $\xi[F]=[k]$; the sets $C_x$ are clopen sets, sets that are both closed and open.)

