Prove that any 3-regular graph with diameter 2 and order 10 is isomorphic to Petersen graph. As I was reading this paper [Classification of Regular Planar Graphs with Diameter Two, DOI: 10.1007/s10114-005-0607-4], I noticed a lemma that cites in Algebraic Graph Theory.

Case 3: Let $p=10$.  It is well-known that any $3$-regular graph with diameter $2$ and order $10$ is isomorphic to the Petersen graph [....]

[2] Biggs, N.: Algebraic Graph Theory, Second Edition, Cambridge University Press, Cambridge, 1993
But I looked up reference 2, which is the algebraic graph theory book, and unfortunately I didn't find any narrative pointing to this lemma.
I tried to prove it, but it didn't seem easy.
 A: Let $G$ be a graph with the stated properties.  Let's label one of the vertices $0$.  It has three neighbors, so let's label them $1$, $2$, and $3$ in some order.  Then we will label the other two neighbors of $1$ with $11$ and $12$ in some order, and so on.  So now as much of the graph as we know about looks like this.

Since $G$ has ten vertices, we have accounted for all of them.  Note that it was critical that these vertices are all distinct in order to cover ten vertices with 3-regularity and diameter 2, so there can't be any 3-cycles in G.
We are missing two edges on each of the outside vertices, and we need to connect them in such a way that the diameter of the graph is 2.  Without loss of generality, let's attach 11 to 21 and 32.  Now, 11 is within two steps of every vertex except 22 and 31.  If 22 connected to 21, we'd have a 3-cycle, so it must be that 21 connects to 31 and 32 connects to 22.  The only vertices left for 12 to connect to are 22 and 31, and the $G$ now looks like this.

This is a common diagram of the Petersen graph.  If you need to compare it to the pentagon and pentagram diagram, one 5-cycle is (0, 2, 22, 32, 31) and the other is (1, 11, 21, 31, 12).
