If a graph is planar, how to prove any of its minors is also planar? In my head this is trivial, but I got this question as part of an assignment, so I am pretty sure my argumentation has to be more complete.
If a graph is planar, then removing any edge will keep it planar, that is trivial, removing any vertex and its adjacent edges will keep it planar, also trivial, and removing two vertices a, b and an edge e = (ab), by replacing them with a single vertex v, and appending all the edges incident to a and b to v will still keep it planar (contraction), because we are not creating any new edges, just extending some to an already existing point.
Then, since a minor can be formed by applying any of these 3 processes, any minor in G will still be planar. 
However, I may be wrong here, how could I write a more adequate proof?
I can't use Kuratowski's theorem for this proof. 

Suppose that $H$ a minor of $G$ is not planar. Let $\{G_1,\ldots,G_l \}$ be the set of all the possible graphs obtained from $G$ by any number of edge contractions. Note that since $G$ is planar then $G_i$ is also planar $\forall i \in \{1,\ldots,l\}$.
Since $H$ is minor of $G$, then $H$ is subgraph of $G_k$, for some $k \in \{1,\ldots,l\}$. Since $H$ is not planar, then $G_k$ is also not planar, contradiction.
