New to proofs, need help with how to approach a beads-and-wires proof puzzle I am working on a proof puzzle about beads and wires. We are given 4 axioms about the objects you can create with beads and wires.


*

*Axiom 1. You must have exactly 3 beads.

*Axiom 2. There is exactly one wire between each pair of beads.

*Axiom 3. Not all beads can be on the same wire.

*Axiom 4. Any pair of wires has at least one bead in common.


We are asked to prove the following theorem: 

Theorem. No bead can be on all wires for all possible bead-wire models. 

I am new to proofs and have only had some experience doing simple proofs based on real numbers. Therefore, I am having issues on how to rigorously use these axioms about beads and wires in a way of proving the theorem. 

So far, I can see that if one bead is on all wires, that if you follow Axiom 2 and Axiom 4 that you would need more that 4 beads if you don’t have all beads on one wire (contradicting Axiom 1). 

I am just having issues of how to rigorously represent these objects and make that into a proof instead of just a visual intuition.
Any help with how to get started would be much appreciated!

Sketch:
Label the beads $B_1,B_2,B_3$ and suppose that $B_1$ was on all the wires, we will derive a contradiction.  
Let $W_{ij}$ be the unique wire containing both $B_i,B_j$, for $i\neq j$. 
If $W_{12},W_{23}$ were the same wire then all three beads would be on that one wire, which would contradict Axiom $3$.  Thus those two wires are different. 
But if $B_1$ were on all the wires then it would have to be on $W_{23}$.  Hence $W_{23}$ connects $B_1$ and $B_2$ so, by Axiom $2$ we must have $W_{12}=W_{23}$.
I don't believe Axiom $4$ is needed in this proof.  
