Affirmed.
I offer the following (hopefully constructive) criticisms:
- You should recognise why this graph drawing is important, e.g. by writing:
"We can see that the original graph has a subgraph homeomorphic to $K_{3,3}$ by redrawing it as follows: [[blah blah blah]]. Therefore it's not planar by Kuratowski's Theorem."
- There's (arguably) a better way to draw the second graph; see below:


These were drawn using TikZ. Here's the LaTeX code:
\begin{tikzpicture}
\tikzstyle{vertex}=[circle,draw=black,fill=black!20,minimum size=18pt,inner sep=0pt]
\draw (0,0) node (c){};
\draw (c)++(0*360/4+135:3) node[vertex] (A){$A$};
\draw (c)++(1*360/4+135:3) node[vertex] (C){$C$};
\draw (c)++(2*360/4+135:3) node[vertex] (D){$D$};
\draw (c)++(3*360/4+135:3) node[vertex] (B){$B$};
\draw (c)++(0*360/4+135:1.5) node[vertex] (E){$E$};
\draw (c)++(1*360/4+135:1.5) node[vertex] (G){$G$};
\draw (c)++(2*360/4+135:1.5) node[vertex] (H){$H$};
\draw (c)++(3*360/4+135:1.5) node[vertex] (F){$F$};
\draw[ultra thick] (A) -- (B) -- (D) -- (C) -- (A);
\draw[ultra thick] (A) -- (E) -- (H) -- (D);
\draw[ultra thick] (B) -- (F) -- (G) -- (C);
\draw[ultra thick] (E) -- (G);
\draw (F) -- (H);
\end{tikzpicture}
and for the second drawing:
\begin{tikzpicture}
\tikzstyle{vertex}=[circle,draw=black,fill=black!20,minimum size=18pt,inner sep=0pt]
\draw (0,0) node (c){};
\draw (c)++(0,3)++(2,0) node[vertex] (G){$G$};
\draw (c)++(0,3) node[vertex] (A){$A$};
\draw (c)++(0,3)++(-2,0) node[vertex] (D){$D$};
\draw (c)++(-2,0) node[vertex] (E){$E$};
\draw (c) node[vertex] (C){$C$};
\draw (c)++(2,0) node[vertex] (B){$B$};
\draw (c)++(-2,1.5) node[vertex] (H){$H$};
\draw (c)++(2,1.5) node[vertex] (F){$F$};
\draw[ultra thick] (A) -- (B) -- (D) -- (C) -- (A);
\draw[ultra thick] (A) -- (E) -- (H) -- (D);
\draw[ultra thick] (B) -- (F) -- (G) -- (C);
\draw[ultra thick] (E) -- (G);
\draw (F) -- (H);
\end{tikzpicture}