I was wondering about the title question:
What is the fewest vertices a non-planar subgraph of the Knight's graph can have?
I have found a subgraph with 14 10 vertices that is non-planar
Contract the blue edges to get $K_{3,3}$
And it is pretty easy to show that any subgraph of the knight's graph of size 7 is planar. I just used a proof by exhaustion, showing that you can't make $K_5$ in two contractions and you can't make $K_{3,3}$ in one. I will leave the proof out of this question for the sake of brevity.