Every three-connected graph with no $K_5$ or $K_{3,3}$ minor is planar

A three-connected graph of order four is $K_4$, which has the usual plane drawing. Assume the result for smaller three-connected graphs that exclude the two minors.

If $|G|>4$, choose $xy$ from [L1]. The contraction $H=G/xy$ remains three-connected and excludes the forbidden minors because the minor relation is transitive. It has a plane drawing by the induction hypothesis.

Let $v$ be the contracted vertex and delete it from the plane drawing of $H$. Its incident faces merge into one face of $H-v$, and every former neighbour of $v$ lies on that face boundary. Since $H-v$ is two-connected, the boundary is a cycle $C$. Put $X=N_G(x)\setminus\{y\}$ and $Y=N_G(y)\setminus\{x\}$, both viewed on $C$. Enumerate $X$ cyclically and let the intervening $X$-arcs partition $C$.

All vertices of $Y$ lie on one intervening $X$-arc. Otherwise two vertices of $Y$ alternate on $C$ with two vertices of $X$; the paths through $x$ and $y$, together with the two arcs of $C$, form a subdivision of $K_{3,3}$. The only remaining alternating degeneracy is three common neighbours of $x,y$, which together with $x,y$ form a subdivision of $K_5$. Both contradict [L2].

Delete from the drawing of $H$ the edges at $v$ corresponding only to $Y\setminus X$, and regard $v$ as $x$. The arc of $C$ containing $Y$ bounds, together with the two adjacent $x$-edges, a face of this drawing by polygonal separation and face containment. Place $y$ in that face, join it polygonally to all vertices of $Y$, and draw $xy$ inside a small neighbourhood of $x$. The arcs have disjoint interiors and recover a plane drawing of $G$.

The base and contraction step prove the lemma for all finite orders.