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'Hamiltonian Cube' printed from https://nrich.maths.org/
The image above shows a possible path. Each edge joining a corner
to a face centre has length $\frac{1}{\sqrt{2}}$ (by Pythagoras'
Theorem), while each edge which joins two adjacent corners has
length $1$. So the length of the path above is $1+6\sqrt{2}$. This
is the length of the shortest path to pass through all the
vertices.
To prove this, note the length of the shortest path must be at
least $\frac{13}{\sqrt{2}}$. Such a path would move alternately
between corners and faces, but as there are $8$ corners and only
$6$ faces, so this is impossible. So at least one of the edges must
join to corners, and so the shortest length is $1+6\sqrt{2}$.