Why is the minimal uncountable well-ordered set $S_{\Omega}$ unique? In section 10 of Topology by Munkres, the minimal uncountable well-ordered set $S_{\Omega}$ is introduced. Furthermore, it is remarked that,


  Note that $S_{\Omega}$ is an uncountable well-ordered set every section of which is countable. Its order type is in fact uniquely determined by this condition.


However, how to justify its uniqueness?
This stems from the fact that given two well-ordered sets $X, Y$ exactly one of the following holds (see this previous question):


$X$ is order-isomorphic to $Y$;
There is a (unique) $a \in X$ such that $\{ x \in X : x &lt; a \}$ is order-isomorphic to $Y$; or
There is a (unique) $b \in B$ such that $\{ y \in Y : y &lt; b \}$ is order-isomorphic to $X$.


So suppose $X$ is another well-ordered set which is not order-isomorphic to $S_\Omega$.  Then either


there is a (unique) $b \in S_\Omega$ such that $\{ y \in S_\Omega : y &lt; b \}$ is order-isomorphic to $X$, and since  $\{ y \in S_\Omega : y &lt; b \}$ is countable, it follows that $X$ is also countable.
there is a (unique) $a \in X$ such that  $\{ x \in X : x &lt; a \}$ is order-isomorphic to $S_\Omega$.  But then $\{ x \in X : x &lt;_X a \}$ is itself an initial section of $X$ which is uncountable.


So any well-ordered set which is not order-isomorphic to $S_\Omega$ is either countable, or has an uncountable proper initial section.
Suppose that $\alpha$ and $\beta$ are two uncountable ordinals.  Suppose further that for every $\gamma&lt;\alpha$, $\gamma$ is countable.  Likewise, suppose that for very $\eta&lt;\beta$, $\eta$ is countable.  You want to show that $\alpha=\beta$. 

Suppose that $\alpha&lt;\beta$.  Then $\eta=\alpha$ is uncountable, and satisfies $\eta&lt;\beta$.  This is a contradiction.  Thus $\alpha\not&lt;\beta$.  Likewise $\beta\not&lt;\alpha$.