a) Show that $K_{3,3}-e$ is flat for all $e \in E(K_{3,3})$.
b) Show that $K_{5}-e$ is planar for all $e \in E(K_{5})$.
Attempt:
a) Since we have removed an edge from $K_{3,3}$, it follows that the new graph does not have $K_{3,3}$ as a subgraph. By Kuratowski's theorem, $K_{3,3}-e$ is planar.
b) Since we have removed an edge of $K_5$, it follows that the new graph does not have $K_5$ as a subgraph. By Kuratowski's theorem, $K_5-e$ is planar.
On
First, I would say you're better off using Wagner's theorem instead of Kuratowski because Wagner just considers minors, which are easier to show for small graphs such as this.
Second, Wagner/Kuratowski both say that your graph can contain neither $K_{3,3}$ nor $K_5$. So the harder part is not to show that $K_{3,3}-e$ does not contain $K_{3,3}$, but also that it does not contain $K_{5}$. Same for the other.
There are many ways to do this, but the general idea is to find some property of $K_5$ that is not there in $K_{3,3}$ (and therefore not in $K_{3,3}-e$), and vice versa. For example, the latter does not have any vertex of degree $4$, but all vertices of $K_5$ have degree $4$. For the other, simply look at the number of vertices.
Be careful about the kind of property you choose. The strategy I showed implicitly assumed a subgraph-closed property, aka if it is true for $G$, it is true for all subgraphs of $G$. This is not necessarily true (trivial example: if $G$ has an odd number of edges, then there is clearly some subgraph of $G$ with an even number of edges). If you do pick a property that is not subgraph closed, then you have to show it holds for $K_{3,3}-e$ specifically as well.
I think the natural way to prove these specific graphs are planar is to provide an embedding. When trying to give embedding for graphs it usually helps to place the first $3$ of them in a triangle in the outside, and then triangulate when possible. In the case of bipartite graphs you can't have triangles but you can try to get as many quadrilateral faces as possible.