Prove that this infinite planar map can still be colored with four colors. Here is quite challenging problem from Enderton's popular textbook A Mathematical Introduction to Logic.

In 1977 it was proved that every planar map can be colored with four colors. Of course, the definition of "map" requires that there be only finitely many countries. But extending the concept, suppose we have an infinite (but countable) planar map with countries $C_1, C_2, C_3, ...$. Prove that this infinite planar map can still be colored with four colors. (Suggestion: Partition the sentence symbols into four parts. One sentence symbol, for example, can be used to translate, "Country $C_7$ is colored red." Form a set $\Sigma_1$ of wffs that say, for example, $C_7$ is exactly one of the colors. Form another set $\Sigma_2$ of wffs that say, for each pair of adjacent countries, that they are not the same color. Apply compactness to $\Sigma_1 \cup \Sigma _2$

Since I'm not much of the math logician, it's difficult for me to make a rigorous proof.  My attempt would be to first check the satisfiability of every finite subset of $\Sigma_1$ and $\Sigma_2$.  Then, take the union of those sets and finally apply compactness theorem, which states that

A set of wffs is satisfiable iff every finite subset is satisfiable.

The thing is: I wonder how the proof goes since Enderton's textbook is sometimes brief in some topics and the way he proves theorems and examples.  He skips steps to the readers, assuming that they can prove them by themselves.  Thus, I am a bit lost of how I should prove this.
Any suggestions or advices or comments?
 A: Consider the following set of propositional variables $\{P_{n,i}:1\leq i\leq4\ \wedge\ n\in\mathbb N\}$. We are interpreting $P_{n,i}$ as the $n$-th country has colour $i$. Let $\Sigma$ be the following set of sentences:
$1$. $P_{n,1}\vee P_{n,2}\vee P_{n,3}\vee P_{n,4}$ for all $n\in\mathbb N$,
$2$. $\neg(P_{n,i}\wedge P_{n,j})$ for all $1\leq i<j\leq4$ and $n\in\mathbb N$,
$3$. $\neg(P_{n,i}\wedge P_{m,i})$ for all $1\leq i\leq 4$ and all pair of adjacent countries $C_n$ and $C_m$.


*

*says that every country gets a colour, 

*says that each country gets at most one colour and,

*says that no two adjacent countries get the same cloud.


$\Sigma$ is finitely satisfiable by hypothesis, so by compactness, is satisfiable.
Any truth valuation witnessing gives you the decided colouring.  
A: We need to define what it means for an infinite graph to be planar (note that we did not say countable). We will cheat and say that such a graph $G$ is planar iff every finite subgraph of $G$ is planar.
Take the usual language of graph theory, using a single binary predicate symbol $E(x,y)$ that says $x$ and $y$ are joined by an edge. Add to it four unary predicate symbols $C_1,C_2,C_3,C_4$, with axioms that say that for every vertex $x$, precisely one of the $C_i(x)$ holds. Add to that sentences that say that if $E(x,y)$, then it is not the case that $C_i(x)$ and $C_i(y)$ both hold ($i=1,2,3,4$).
Let the resulting theory be $T$.
Let $G$ be an an infinite graph. Add to the language of $T$ a constant symbol $a_v$ for every vertex $v$ of $G$.  Add to the axioms of $T$ the sentences $E(a_u,a_v)$ for all $u$ and $v$ that are joined by an edge of $G$, and $\lnot E(a_u,a_v)$ for all $u$, $v$ that are not joined by an edge of $G$. (So we are adding to $T$ the diagram of $G$.) Call the resulting theory $T_G$.
Any finite subset of the axioms of $T_G$ is consistent, since every finite planar graph is $\le 4$-colourable. Thus by Compactness $T_G$ has a model $G'$. The interpretations of the $C_i$ give a colouring of $G'$ in four colours or less.
The graph $G$ is a subgraph of $G'$, and the colouring of $G'$ in four colours or less restricts to a $\le 4$-colouring of $G$.
