Suppose $G$ is a nonplanar graph of order $7.$ By Kuratowski's theorem, $G$ has a subgraph $H$ which is either a subdivision of $K_5$ or a subdivision of $K_{3,3}.$ A subdivision of $K_5$ has $5$ vertices of degree $4$; a subdivision of $K_{3,3}$ has $6$ vertices of degree $3.$ If $H$ is a subdivision of $K_5,$ then $\bar G$ has at least $5$ vertices of degree $\le2,$ and therefore (by Kuratowski's theorem) must be planar. Therefore, we may assume that $H$ is a subdivision of $K_{3,3}.$ Since $G$ has only $7$ vertices, either $H=K_{3,3},$ or else $H$ is obtained from $K_{3,3}$ by replacing one edge of $K_{3,3}$ with a path of length $2.$ In either case it is easy to check that the graph $K_7-E(H)$ (and therefore its subgraph $\bar G$) is planar.
On the other hand, the graph $G=K_{3,3}+2K_1$ is a nonplanar graph of order $8$ whose independence number is $5$; thus both $G$ and $\bar G$ are nonplanar graphs.
Take an $8$-cycle $v_1v_2v_3v_4v_5v_6v_7v_8v_1$ and add two more edges $v_1v_5$ and $v_2v_6.$ The resulting graph $G$ is clearly planar. I leave it to you to verify that the complement $\overline G$ is also planar. ($\overline G$ is a cube with $6$ additional edges, namely, a diagonal drawn on each face.)
Methodology? It just seemed like a good idea to start with a maximal planar graph; the more edges in the graph, the fewer edges in the complement, and the easier it will be to draw the complement in the plane. So I took a cube (no particular reason, just because it's a nice planar graph with $8$ vertices) and triangulated each of the faces. The complement turned out to be a cycle plus two diagonals, which was planar and easy to describe.
P.S. To see that $G$ is planar, let
$v_1=(0,0),\ v_2=(1,1),\ v_3=(1,2),\ v_4=(1,3),$
$v_5=(1,4),\ v_6 (2,4),\ v_7=(2,2),\ v_8=(2,0),$ and draw the edges $v_1v_2,\ v_2v_3,\ v_3v_4,\ v_4v_5,\ v_5v_6,\ v_6v_7,\ v_7v_8,\ v_8v_1,\ v_1v_5,\ v_2v_6$ as straight line segments.
Best Answer
It follows from the Euler's formula that a simple planar graph $G$ with $m$ edges and $n\geq 3$ vertices must satisfy (see here) $$\tag{1}m\leq 3n-6.$$ For a graph $G$ with $m$ edges and $n$ vertices, its complement $\overline{G}$ has $\displaystyle\frac{n(n-1)}{2}-m$ edges. Therefore, if $\overline{G}$ is also planar, by $(1)$ we have $$\tag{2}\frac{n(n-1)}{2}-m\leq 3n-6.$$ Adding $(1)$ and $(2)$, we obtain $$\frac{n(n-1)}{2}\leq 6n-12,$$ which implies that $n\leq 10$.