Four Color Theorem for Finite Maps implies Four Color Theorem for Infinite Maps

Theorem
Suppose that any finite planar graph can be assigned a proper vertex $k$-coloring such that $k \le 4$.

Then the same is true of any infinite planar graph.

Proof
Let $G$ be an infinite planar graph.

Let $C$ be a set of faces of $G$.

For each $c \in C$ let $p_c^1, p_c^2, p_c^3, p_c^4$ be propositional symbols, where
 * $p_c^i$ is true iff the color of face $c$ is $i$.

Let $\mathcal P_0$ be the vocabulary define as:
 * $\mathcal P_0 = \left\{{p_c^1, p_c^2, p_c^3, p_c^4: c \in C}\right\}$

Let $\mathbf H$ be the set of all statements with the following forms:


 * $(1): \quad$ $p_c^1 \lor p_c^2 \lor p_c^3 \lor p_c^4$ for each $c \in C$
 * $(2): \quad$ $p_c^i \implies \neg p_c^j$ for each $c \in C$ and for each $i \ne j$
 * $(3): \quad$ $\neg \left({p_c^i \land \neg p_{c'}^i}\right)$ for each $i$ where $c$ and $c'$ are adjacent.

Let $\mathcal M$ be a model for $\mathbf H$ which corresponds to a proper vertex $4$-coloring of $G$.

By hypothesis, any finite planar graph can be assigned a proper vertex $4$-coloring.

Thus every finite subgraph of $G$ has a proper vertex $4$-coloring.

That is, every finite subset of $\mathbf H$ has a model.

By the Compactness Theorem of Propositional Calculus, $\mathbf H$ has a model.

Hence the entirety of $G$ can be assigned a proper vertex $k$-coloring.