I won't do the given problem, but I'll do one that may make it clearer how one can proceed. Consider the graph given by the vertices and edges of a cube, together with one body diagonal (that is, one extra edge, passing through the center of the cube, and joining two opposite vertices of the cube). Let's prove this isn't planar by finding a subgraph isomorphic to $K_{3,3}$.
First of all, without that body diagonal, the graph is planar, so our subgraph must include that edge. Let's call its endpoints $A$ and $X$. There are three other vertices adjacent to $A$; pick two of them (by symmetry, it doesn't matter which two) and call them $Y$ and $Z$. So at this point, we have $A$ adjacent to $X$, $Y$, and $Z$.
Now there is another vertex adjacent to all three of $X$, $Y$, and $Z$; call it $B$.
Now there's a vertex, as yet unused, adjacent to both $X$ and $Z$; call it $C$. $C$ is not adjacent to $Y$, but there is a path running from $C$ to $Y$ via vertices we haven't used for anything else. So that gives you a subgraph in which there are independent paths joining each of $A,B,C$ to each of $X,Y,Z$; in other words, a subgraph homeomorphic to $K_{3,3}$; in other words, a proof of nonplanarity of "cube plus body diagonal".