Proof: By Euclid

Let the diameter $AB$ of the given sphere be laid out, and let it have been cut in half at $C$.
And let the semicircle $ADB$ have been drawn on $AB$.
And let $CD$ be drawn from $C$ at right angles to $AB$.
And let $DB$ have been joined.
And let the square $EFGH$, having each of its sides equal to $DB$, be laid out.
And let $HF$ and $EG$ have been joined.
And let the straight line $KL$ have been set up, at point $K$, at right angles to the plane of square $EFGH$ [Prop. 11.12].
And let it have been drawn across on the other side of the plane, like $KM$.
And let $KL$ and $KM$, equal to one of $EK$, $FK$, $GK$, and $HK$, have been cut off from $KL$ and $KM$, respectively.
And let $LE$, $LF$, $LG$, $LH$, $ME$, $MF$, $MG$, and $MH$ have been joined.
And since $KE$ is equal to $KH$, and angle $EKH$ is a right angle, the (square) on the $HE$ is thus double the (square) on $EK$ [Prop. 1.47].
Again, since $LK$ is equal to $KE$, and angle $LKE$ is a right angle, the (square) on $EL$ is thus double the (square) on $EK$ [Prop. 1.47].
And the (square) on $HE$ was also shown (to be) double the (square) on $EK$.
Thus, the (square) on $LE$ is equal to the (square) on $EH$.
Thus, $LE$ is equal to $EH$.
So, for the same (reasons), $LH$ is also equal to $HE$.
triangle $LEH$ is thus equilateral.
So, similarly, we can show that each of the remaining triangles, whose bases are the sides of the square $EFGH$, and apexes the points $L$ and $M$, are equilateral.
Thus, an octahedron contained by eight equilateral triangles has been constructed.
So, it is also necessary to enclose it by the given sphere, and to show that the square on the diameter of the sphere is double the (square) on the side of the octahedron.
For since the three (straight lines) $LK$, $KM$, and $KE$ are equal to one another, the semicircle drawn on $LM$ will thus also pass through $E$.
And, for the same (reasons), if $LM$ remains (fixed), and the semicircle is carried around, and again established at the same (position) from which it began to be moved, then it will also pass through points $F$, $G$, and $H$, and the octahedron will have been enclosed by a sphere.
So, I say that (it is) also (enclosed) by the given (sphere).
For since $LK$ is equal to $KM$, and $KE$ (is) common, and they contain right angles, the base $LE$ is thus equal to the base $EM$ [Prop. 1.4].
And since angle $LEM$ is a right angle - for (it is) in a semicircle [Prop. 3.31] - the (square) on $LM$ is thus double the (square) on $LE$ [Prop. 1.47].
Again, since $AC$ is equal to $CB$, $AB$ is double $BC$.
And as $AB$ (is) to $BC$, so the (square) on $AB$ (is) to the (square) on $BD$ [Prop. 6.8], [Def. 5.9] .
Thus, the (square) on $AB$ is double the (square) on $BD$.
And the (square) on $LM$ was also shown (to be) double the (square) on $LE$.
And the (square) on $DB$ is equal to the (square) on $LE$.
For $EH$ was made equal to $DB$.
Thus, the (square) on $AB$ (is) also equal to the (square) on $LM$.
Thus, $AB$ (is) equal to $LM$.
And $AB$ is the diameter of the given sphere.
Thus, $LM$ is equal to the diameter of the given sphere.
Thus, the octahedron has been enclosed by the given sphere, and it has been simultaneously proved that the square on the diameter of the sphere is double the (square) on the side of the octahedron.
(Which is) the very thing it was required to show.
∎

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References

Adapted from (subject to copyright, with kind permission)

Fitzpatrick, Richard: Euclid's "Elements of Geometry"

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Proof: By Euclid

References

Adapted from (subject to copyright, with kind permission)