TITLE: Rado graph containing infinitely many isomorphic subgraphs 
QUESTION [5 upvotes]: The Rado graph contains every finite graph as an induced subgraph. It surely contains some finite graphs infinitely often as an induced subgraph, e.g. $K_2$. Does it contain all finite graphs infinitely often as an induced subgraph? Or can an example of a graph be given that is not contained infinitely often?

REPLY [3 votes]: I realise that the question is almost three years old, but maybe that's not that long in Maths. 
I am not sure who the following is originally due to, but as far as I am aware, it's the standard to show that the Rado graph contains every countable graph. The idea is to start with whatever graph you want and construct the Rado graph around it.
Let $G=G_0$ be any countable graph. For $i>0$ define a new Graph $G_i$ as follows: 

The vertices $V(G_{i})$ of $G_i$ are all of $V(G_{i-1})$ plus an extra vertex $v_A$ for every finite subset $A$ of $V(G_{i-1})$.
All edges of $V(G_{i-1})$ are also edges of $V(G_{i})$
Add an edge between $a\in V(G_{i-1})$ and $v_{A}\in V(G_{i})$ whenever $a\in A$.

Let $R$ be the union of the graphs $G_i$ over all $i>0$. Then show that $R$ is the Rado graph. Clearly, $R$ contains $G=G_0$.
One of the nice things about this construction is that you can show without much difficulty that any automorphism of $G$ extends to an automorphism of $R$. So not only does the Rado graph contain every countable graph $G$, but it contains "special" copies of $G$ with the above extension property.