The Five Color Theorem

Base case. If $G$ has at most 5 vertices, color each vertex a different color; at most 5 colors are used, so the claim holds.

Inductive step, setup. Assume every planar graph with fewer than $n$ vertices is 5-colorable, and let $G$ be planar with $n$ vertices. Every simple planar graph satisfies $E \le 3V - 6$ (edge bound proved via Euler's formula), so the sum of all degrees is at most $2(3n-6) = 6n-12$, which is less than $6n$. Hence the average degree is less than 6, so some vertex $v$ has degree at most 5.

Remove $v$ to get a smaller planar graph with $n-1$ vertices; by the inductive hypothesis it has a proper 5-coloring. If $v$ has at most 4 neighbors, at most 4 colors are used among them, leaving a free color for $v$, and we are done.

The remaining case is deg$(v) = 5$ and all 5 colors appear, once each, among the 5 neighbors of $v$. List the neighbors in the cyclic order they appear around $v$ in the planar drawing as first, second, third, fourth, fifth, colored 1, 2, 3, 4, 5 respectively. Consider the subgraph $H_{1,3}$ formed by all vertices colored 1 or 3. If the first and third neighbors lie in different connected components of $H_{1,3}$, swap colors 1 and 3 throughout the component containing the first neighbor; this stays a proper coloring (it only touches vertices colored 1 or 3), and now the first neighbor is colored 3, so color 1 is free for $v$.

Otherwise the first and third neighbors lie in the same component of $H_{1,3}$, joined by a path alternating colors 1 and 3. Together with $v$ and the two edges to the first and third neighbors, this path closes a cycle in the plane that separates the second neighbor from the fourth neighbor (by the Jordan curve theorem, since the cyclic order around $v$ is first, second, third, fourth, fifth). Consequently no path of alternating colors 2 and 4 can connect the second and fourth neighbors, since such a path would have to cross the 1-3 cycle. So we swap colors 2 and 4 throughout the component of $H_{2,4}$ containing the second neighbor; this frees color 2 for $v$.

In every case $v$ receives one of the 5 colors without conflicting with any neighbor, extending the coloring of $G$-v to all of $G$. By induction, $\chi(G) \le 5$ for every planar graph.