Simple example of uncountable ordinal

The proof that the set $\Omega$ of all countable ordinals is uncountable is not difficult. First, it's an ordinal. Next, if $\Omega$ were countable, then $\Omega + 1$ would also be a countable ordinal. Finally, it is impossible for $\Omega + 1$ to be in $\Omega$, because that would mean that $\Omega + 1 < \Omega$.

Here is another way of arguing which is slightly different: Consider the set $X={\mathcal P}({\mathbb N}\times{\mathbb N})$ of all subsets of ${\mathbb N}^2$. Given $E\subseteq {\mathbb N}\times{\mathbb N}$, let $A_E$ be the set of all numbers that appear in either the domain or range of $E$ (this is sometimes called the field of $E$). If it happens that $E$ is a well-ordering of $A_E$, let $\alpha_E$ be the unique ordinal that is order isomorphic to $(A_E,E)$. Otherwise, let $\alpha_E=0$. Then $\{\alpha_E\mid E\in X\}$ is a set (is the image of $X$ under the map $E\mapsto\alpha_E$) and it is obvious that it consists precisely of the countable ordinals. One easily sees then that it is itself an ordinal, and uncountable (since no ordinal belongs to itself). This is $\omega_1$.

The only two that I think have simple proofs are $\Omega$, as Carl says, and $2^{\aleph_0}$, the set of infinite binary sequences. Cantor's diagonal argument proves this one uncountable. Alternately, as all cardinals are ordinals, any cardinal greater than $\aleph_0$ is an uncountable ordinal just from the definition of countability. That's simple to write, but probably not what you are looking for.