Gleason's article "Angle Trisection, the Heptagon, and the Triskaidecagon" (also available here) mentions a construction due to Plemelj:
Draw the circle with center $O$ passing through $A$ and on it find $M$ so that $AM=OA$. Bisect $OM$ at $N$, and trisect at $P$, and find $T$ on $NP$ so that $\angle NAT=\frac13\angle NAP$. $AT$ is the needed side of the heptagon.
To validate Plemelj's construction, we must prove that $AT=OA\left(2\sin\dfrac{\pi}{7}\right)$. Since $2\cos\dfrac{2\pi}{7}=2-\left(2\sin\dfrac{\pi}{7}\right)^2$, it follows from $x=2\cos\dfrac{2\pi}{7}$ being a root of $x^3+x^2-2x-1=0$ that $2\sin\dfrac{\pi}{7}$ is a root of
$$(2-x^2)^3+(2-x^2)^2-2(2-x^2)-1=0$$
the other roots being $-2\sin\dfrac{\pi}{7}$, $\pm 2\sin\dfrac{2\pi}{7}$, and $\pm 2\sin\dfrac{3\pi}{7}$. The equation can be factored as
$$\left(x^3+\sqrt 7\left(x^2-1\right)\right)\left(x^3-\sqrt 7\left(x^2-1\right)\right)=0$$
The roots corresponding to the first cubic factor are $2\sin\dfrac{\pi}{7}$, $-2\sin\dfrac{2\pi}{7}$, and $-2\sin\dfrac{3\pi}{7}$. Writing the first cubic factor in the form
$$\left(\frac1{x}\right)^3-\frac1{x}=\frac1{\sqrt 7}$$
and making the substitution $\dfrac1{x}=\dfrac2{\sqrt 3}\cos\,\psi$ yields the equation $\cos\,3\psi=\sqrt{\dfrac{27}{28}}$. The desired root corresponds to the choice
$$\psi=\frac13\arccos\sqrt{\frac{27}{28}}=\frac13\arctan\frac1{3\sqrt 3}$$
which yields
$$2\sin\frac{\pi}{7}\cos\,\psi=\frac{\sqrt 3}{2}$$
From the figure, we have $\angle NAP=\arctan\dfrac1{3\sqrt 3}$, so $\angle NAT=\psi$. We thus have $AT\cos\,\psi=AN=\dfrac{\sqrt 3}{2}OA$, and from this we also have $AT=OA\left(2\sin\dfrac{\pi}{7}\right)$.
