Our situation: We have the assertion we want to prove, let us call it $P_A$ , depending on a set $A$. We know already $A$ is true for all finite $A$ and now want to show it for all infinite sets, too.
[If I understood correctly you are not inetrested in further details for the proof of finite sets, so I say we know it for finite sets.]

A $k$-coloring of $\mathbb{R}$ is a map from $\mathbb{R}$ to $\lbrace 1, \dots, k \rbrace$ (or whatever $k$ element set). The toplogogy on the set of colors will be the discrete topology so every set is open (and closed). Since the set is finite this is compact.

The set of all colorings is, by defnition, $K = \lbrace 1, \dots, k \rbrace^{\mathbb{R}}$ .
As a product of compact sets this is compact with respect to the product/Tychonoff toplogy, as you say.

Now, for some $x$ you denote $K_x$ the subset of $K$ such that $x+ S$ is colorful in this coloring; let us say such a coloring 'works for $x$'.

Now for a set $X$ to show $P_X$ we need to show there is some coloring such that $x+S$ is colorful for each $x \in X$. In other words, we need a coloring that works for each $x \in X$, which is nothing but saying the intersection of $K_x$ over $x \in X$ is nonempty.

Since we know $P_A$ for finite $A$ we know that the intersection of $K_x$ for $x \in A$ is nonempty.

Suppose for a moment we know that $K_x$ is closed. Then the complements are open. Now, if we find an infinite set $X$ such that the intersection $K_x$ over $x \in X$ is empty, we can say equivalently the union of the complements (open sets!) is the full space, and by compactness, we get that there is a finite subunion that already convers the full space. Yet, then the finite intersection of the $K_x$ corresponding to this subunion would be empty. A contradition to the fact we know the result for finite sets.

So what remains to assert is that $K_x$ is closed, for each $x$. To see this perhaps it is easier to show that the complement is open (if only as the open sets of product topology are possibly more familiar). So, we show the set of all colorings that do not work for $x$ is open.

Now we note that whether a coloring works or does not work for some $x$ depends only on the values it has for the elements of $x+S$, a *finite* set. For every other point we can have any value.
So the set of colorings that do not work is certainly of the form
$$ N \times \prod_{y \notin x+S} \lbrace 1, \dots, k \rbrace $$
where $N$ is a subset of $\prod_{y \in x+S} \lbrace 1, \dots, k \rbrace $.
Yet, since $x+S$ is *finite* the space $\prod_{y \in x+S} \lbrace 1, \dots, k \rbrace $ is also finite and the topology is the discrete topology, so that $N$ is certainly open (like any set).
So that
$$ N \times \prod_{y \notin x+S} \lbrace 1, \dots, k \rbrace $$ will be open whatever the $N$.

Thus we just showed that the complement of $K_x$ is open and so $K_x$ ` is closed, completing the argument.

Hope this is about what you were looking for.