I think the configuration we are looking at is
a Sierpinski-gasket-like thing. Call it $T$. Then $T$ consists of three half-size copies of $T$ (I mean actually literally shrunk, so in particular all the inductors are half the physical size -- so if they are coils of wire, they will have half the inductance because inductance scales like area/length) connected together via three 1-henry inductors in the "obvious" places. And then the question is: what is the inductance between two corners of $T$?
Here, courtesy of the OP, is a diagram showing how this works:
being a three-terminal circuit built out of impedances, $T$ is equivalent to some other circuit where each of the three terminals is connected via a two-terminal blob $U$ to a central node. $U$ is the same for all three of them, by symmetry. If the corner-to-corner inductance of $T$ is $L$ then the inductance of $U$ must be $L/2$. (This is essentially the "Lemma".) So the inductance of $U$ in a scaled-down copy of $T$ is, because of the scaling, half of this, or $L/4$.
So diagram 1 is obtained by
taking $T$ (with corner-to-corner inductance $L$), exposing its three half-size $T$s (each with corner-to-corner inductance $L/2$), and turning each of them into three half-size $U$s (each with inductance $L/4$).
Then diagram 2
just adds up the inductances along the resulting "inner" paths. Diagram 3 does the usual star-delta thing to give us an equivalent circuit that's easier to work with. Diagram 4 then works out what the corner-to-corner inductance of the resulting thing has to be. Equating this to $L$ gives us the equations at bottom right, and hence enables us to determine the value of $L$.