Other ways of proving that the set of all countable ordinals is uncountable I know that the standard way of proving that the set of all countable ordinals is uncountable is by stating that if the set is countable, then it incurs Burali-Forti paradox.
Is there other ways of proving this?
 A: I think that all proofs will be similar in some way to the Burali-Forti paradox.
Here is a proof that is slightly different than other proofs on this site, so maybe someone will find it useful.
Note that the Axiom of Regularity is not used anywhere.
Definition: An ordinal is a transitive set well-ordered by $\in$.
Fact: The class of ordinals is transitive and well-ordered by $\in$.
Definition: $\omega_1$ is the class of countable ordinals.
Fact: $\omega_1$ is a set.
To see that $\omega_1$ is an ordinal using the above facts, it remains to observe that
every element of a countable ordinal is a subset of that ordinal and is therefore countable also.
Now suppose toward a contradiction that $\omega_1$ is countable.  Then $\omega_1 \in \omega_1$ by definition.  This contradicts the fact that the class of ordinals is well-ordered by $\in$.
A: The set of all countable ordinals is the supremum of all countable ordinals, which is simply their unions. If $\omega_1$ were countable, $\omega_1+1$ would be countable too and hence $\omega_1+1<\omega_1$. Hence $\omega_1\in\omega_1+1\in\omega_1$, which shows that the set $\{\omega_1,\omega_1+1\}$ has no $\in$-minimum, which is impossible since the ordinals are well-ordered by $\in$.  
A: In Principia Mathematica, Whitehead & Russell presented Burali-Forti's paradox in a form that can dispense with the axiom of infinity. They eventually dispelled the paradox by showing that in higher types there are greater ordinals than any to be found in lower types.
The paradox:

Let $\alpha$ be any ordinal number, then the ordinal number for $0_r,
 1_s, 2_r, 3_r, ... \alpha$ in order of magnitude is $\alpha +
 \overset{.}{1}$; thus $\alpha +\overset{.}{1}$ exists, and is greater
   than $\alpha$. But $\alpha$ is similar to the segment of series of ordinals consisting of the predecessors of  $\alpha$, and is therefore less than the ordinal of all ordinals. Hence the ordinal of all ordinals is greater than every
   ordinal, including itself, which is absurd.

In order to dispel the paradox, it is only necessary to make the types explicit. Let $N$ be the series of all ordinals arranged in order of magnitude; when we say "$P$ is well-ordered $\Rightarrow$ $P$ is less than $N$," we mean, by $N$, a series two types above $P$, i.e. $N$'s terms belong to the type of $P$'s ordinal number ($t‘N_0r‘P$). In other words:
$P$ is well-ordered $\Rightarrow$ $P \lt N(P,P)$ 
Whence
$N(P,P)$ is well-ordered $\Rightarrow N(P,P)\lt N\{N(P,P),N(P,P)\}$
Thus, N is no longer less than itself, and the paradox disappears.
Source: Whitehead & Russell, Principia Mathematica V.3. ✳256. Merchant Books, 1913
