Random graphs are not uncountably categorical Is there a simple proof that the theory of random graphs is not $\lambda$-categorical for uncountable $\lambda$?
 A: Whether this counts as a simple proof or not will probably depend on your background. 
An uncountably categorical countable theory is necessarily $\omega$-stable, which the random graph is very far from, which is easy to see directly. Alternatively, you may notice that no nonalgebraic type is stationary (but that requires a characterisation of forking, which does take some effort to establish).
A: I am not sure if this will count as a simple proof, but it is certainly (more) elementary. 

Lemma: Given a graph $G$ of cardinality $\lambda$, there is a graph $H$ which is a model of the theory of the random graph such that $G$ is an induced subgraph of $H$ and $|H|=\lambda$.

Proof of Lemma: We construct $H$ as an increasing union of graphs $\{H_n:n\in\mathbb{N}\}$ constructed inductively. We start with $H_0=(V(G),E(G))$. Supposing now that $H_n$ has been constructed, let $\{A_n^\alpha:\alpha<\kappa\}$ be an enumeration of all the finite subsets of $V(H_n)$. We construct $H_{n+1}$ as follows:


*

*$V(H_{n+1})=V(H_n)\cup\{a_{n+1}^\alpha:\alpha<\kappa\}$, where the elements $a_{n+1}^\alpha$ are new vertices not in $V(H_n)$.

*$E(H_{n+1})=E(H_n)\cup \{(a_{n+1}^\alpha,x):\alpha<\kappa,x\in A_n^\alpha\}$. That is, the new element $a_{n+1}^\alpha$ is connected with all elements in $A_n^\alpha$, and it not connected to the elements in $V(H_n)\setminus A_n^\alpha$. No further edges are added at this stage. 
(In particular, any two new distinct vertices $a_{n+1}^\alpha,a_{n+1}^\beta$ are not connected. This will be important later)


Finally, we take $\displaystyle{H=\bigcup_{n\in\mathbb{N}}H_n}$. 
Note that $G=H_0$ is an induced subgraph of $H$, as the new edges are always between new elements in the construction. Also, $H$ is a model of the random graph, because given finite disjoint subsets $X,Y$ of $V(H)$, there are $n\in\mathbb{N}$ and $\alpha<\kappa$ such $X=A_n^\alpha$ and $Y\subseteq V(H_n)\setminus A_n^\alpha$, and so $a_{n+1}^\alpha\in V(H)$ is connected to every vertex in $X$ and not connected to any vertex in $Y$. $\square$
Now, let $\lambda$ be an arbitrary uncountable cardinal. Let $G_1$ be a complete graph of cardinality $\lambda$, and let $G_2$ be a graph with $\lambda$ many vertices but no edges. Consider the models $H_1,H_2$ of the random graph provided by the construction in the previous Lemma, with $G_1\subseteq H_1$ and $G_2\subseteq H_2$ as induced subgraphs.
Suppose that $f:H_1\to H_2$ is an isomorphism, and let $\{x_\alpha:\alpha<\lambda\}$ be an enumeration of $G_1$. Then, $f(G_1)$ is a complete graph of cardinality $\lambda$ contained in $H_2$. By the construction, $H_2$ is obtained as the union of a countable increasing sequence of graphs $\{H_{2,n}:n\in\mathbb{N}\}$, starting with $H_{2,0}=G_2$. So, $|\{f(x_\alpha):\alpha<\lambda\}\cap V(H_{2,0})|\leq 1$, and by piggeonhole principle there is some $k\in\mathbb{N}$ such that $$|\{f(x_\alpha):\alpha<\lambda\}\cap (V(H_{2,k+1})\setminus V(H_{2,k}))|\geq \aleph_1.$$ 
This is a contradiction, because at every stage there cannot be two new elements that are connected.
