How to explain that the complete graph $K_5$ is non planar without Kuratowski's theorem.

One intuitive argument is as follows: Clearly any $K_3$ will be equivalent to a triangle. (I will hereafter mostly omit "equivalent to" in the interests of brevity.) Now place the fourth point, either inside the triangle, or outside the triangle. Convince yourself that in either case, the results are essentially equivalent: You have a triangle with a point inside it, which is connected to the three vertices of the triangle.

Now, place a fifth point. It is either (a) outside the trisected triangle, or (b) inside one of the trisections. In case (a), the center point cannot access the fifth point, and in case (b), the triangle vertex not attached to the trisection that contains the fifth point cannot access that fifth point. Therefore, $K_5$ is not planar.

This is not by any means a rigorous proof, but it may suffice as an intuitive argument.

If you can use Euler's formula $V-E+F=2$ for plane maps, where $V$ is the number of vertices, $E$ the number of edges, and $F$ the number of faces including the exterior face, then you know that a plane embedding of $K_5$ would have $7$ faces. Each face has (since there are no multiple edges) at least $3$ edges, making at least $21$ edge-face incidences. But there are only $10$ edges, each contributing incidences with just $2$ faces, contradiciton.

Suppose there is a planar embedding of $K_5$.

Claim: each set of three vertices defines a triangular face (possibly the infinite outside face).

Proof: Each set of three vertices must define a triangle, and the two other vertices, being adjacent, must be on the same side of that triangle. The other side must then be a face delimited by the triangle.

Then consider any edge, by the claim it is a side of three different triangular faces (one for each other vertex). But this is clearly impossible, as one side of the edge would be covered twice.