Kuratowski's theorem

Why $K_5$ and $K_{3,3}$ themselves are not planar. $K_5$ has $V=5$ vertices and $E=10$ edges, but the edge bound $E \le 3V - 6$ requires $10 \le 3(5)-6=9$, a contradiction, so $K_5$ is not planar. $K_{3,3}$ is bipartite with $V=6$ vertices and $E=9$ edges; a planar bipartite graph must satisfy the tighter bound $E \le 2V - 4$, requiring $9 \le 2(6)-4=8$, again a contradiction, so $K_{3,3}$ is not planar either.

Subdivisions preserve non-planarity (the easy direction). If a graph $H$ is a subdivision of a graph $G$ (each edge of $G$ replaced by an internally disjoint path), and $G$ is not planar, then $H$ cannot be planar either: any crossing-free drawing of $H$ could be turned into one of $G$ simply by erasing the internal degree-2 vertices on each path and straightening it back into a single edge, which does not introduce any new crossings. So any graph containing a subdivision of $K_5$ or $K_{3,3}$ as a subgraph automatically fails to be planar, which proves the only-if direction of the theorem.

The converse (hard direction), sketch. That every non-planar graph must contain such a subdivision is the deep part, proved independently by Kuratowski in 1930 and, in an equivalent minor-based form, by Wagner in 1937. The argument proceeds by taking a hypothetical minimal non-planar graph containing no subdivision of $K_5$ or $K_{3,3}$ and deriving a contradiction: using Menger's theorem on connectivity, one first reduces to the case where the graph is 3-connected (graphs with a small vertex cut can be split along that cut into smaller planar-or-smaller pieces and reassembled), and then shows directly that every 3-connected non-planar graph of minimum size already contains one of the two forbidden subdivisions. This structural reduction via connectivity is where the real difficulty of the theorem lies.

Conclusion. Combining both directions, a graph is planar exactly when it avoids both forbidden subdivisions, giving the full characterization.