It is a relatively simple exercise to prove that a well-ordered set is order-isomorphic to a subset of $\mathbb R$ (under the usual ordering) if and only if it is countable. You can say that $\mathbb R$ is "too small" to contain any uncountable well-ordered sets.
So my question is, can you embed bigger well-ordered sets in the long line? For those who don't know, the long line can be constructed by taking the minimal uncountable well-ordered set (i.e. $\omega_1$) and taking its Cartesian product with $[0,1)$ under the dictionary order. So obviously $\omega_1$ itself is emebeddable in the long line, just by taking the left endpoints of all the intervals $[0,1)$. But can you embed bigger uncountable ordinals, and if so how big? I'm guessing that you may be able to embed all well-ordered sets with cardinality less than or equal to $\aleph_1$, the cardinality of the set of countable ordinals.
Not that many, really.
The thing about the "long line" is that every proper initial segment is "just a line" (read: an interval of $\Bbb R$). So every proper initial segment can only house countable ordinals.
So no uncountable ordinal greater than $\omega_1$ can be embedded in the long line; and since the long line trivially embeds $\omega_1$, by picking the end-points of the intervals, we get that the only uncountable ordinal that can be embedded in the long line is $\omega_1$ itself.